Neurčitý integrál - Kiwi.mendelu.czuser.mendelu.cz/marik/prez/integraly-cz.pdfNeurčitý...

Neurit integrl

Robert Mak

4. bezna 2012

V tomto souboru jsou vysvtleny a na pkladech s postupnm eenmdemonstrovny zkladn integran metody. Ikonka za integrlem nateintegrl do online aplikace na adresehttp://www.mendelu.cz/user/marik/maw, kter umon vpoet integrlu,vetn automatickch nvrh, jakou metodu zvolit.

cRobert Mak, 2012

http://www.mendelu.cz/user/marik/mawhttp://www.mendelu.cz/user/marik/maw

Obsah

1 Definice neur itho integrlu 5

2 Zkladn vzorce 7

(2x + 3 4x +

6

x3 sinx + ex) dx . . . . . . . . . . . . . . . . . . . . . 8

tgx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

x + 2

x2 + 4x + 5dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

x + 5

x2 + 4dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1

(x + 6)3dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

f (ax + b) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

x + 5

x2 4x + 9dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3 Integrace per-parts 57

cRobert Mak, 2012

(x + 1) lnx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

x sinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

(x 2) sin(2x) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

(x2 + 1) sinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

(x2 + 1)ex dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

x arctgx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

lnx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

ln2 x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

x3 sinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

(x3 + 2x)ex dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

cRobert Mak, 2012

4 Integrace pomoc substituce. 125

sin(lnx)x

dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

xe1x2

dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

x

x4 + 16dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

ex+1

x + 1

dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

tg3 x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

3x + 2 1x + 1

dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

1 +x 1x

dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

5 Dal . . . 192

arcsinx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

cRobert Mak, 2012

1 Definice neur itho integrlu

Definice (neurit integrl, primitivn funkce). Bu I oteven interval, f a Ffunkce definovan na I . Jestlie plat

F (x) = f (x) pro vechna x I , (1)

nazv se funkce F primitivn funkc k funkci f , nebo t neurit integrlfunkce f na intervalu I . Zapisujeme

f (x) dx = F (x).

Existuje-li k funkci f neurit integrl na intervalu I , nazv se funkce f inte-grovateln na I .

Primitivn funkce F (x) je vdy spojit na I , plyne to z existence derivace.

Vta 1 (postaujc podmnka existence neuritho integrlu). Ke kad spojitfunkci existuje neurit integrl.

Definice neuritho integrlu cRobert Mak, 2012

Vta 2 (jednoznanost primitivn funkce). Primitivn funkce je na danm inter-valu k dan funkci urena jednoznan, a na libovolnou aditivn konstantu.Pesnji, plat nsledujc:

Je-li F primitivn funkc k funkci f na intervalu I , plat tot i pro funkci G(x) =F (x) + c, kde c R je libovoln konstanta nezvisl na x.

Jsou-li F a G primitivn funkce k te funkci f na intervalu I , li se ob funkce naintervalu I nejve o aditivn konstantu, tj. existuje c R takov, e

F (x) = G(x) + c pro vechna x I .

Bohuel, ne vdy neurit integrl dokeme efektivn najt. Zatmco problmnalezen derivace funkce sloen z funkc, kter umme derivovat, spovpouze ve sprvn aplikaci vzorc pro derivovn, problm nalzt neuritintegrl i k funkci tak jednoduch, jako je napklad ex

2

je neeiteln ve tdelementrnch funkc.

Definice neuritho integrlu cRobert Mak, 2012

2 Zkladn vzorce

Vta 3. Nech f , g jsou funkce integrovateln na I , c nech je reln slo. Pakna intervalu I plat

f (x) + g(x) dx =

f (x) dx +

g(x) dx,

cf (x) dx = c

f (x) dx.

Vta 4. Nech f je funkce integrovateln na I .

Pak

f (ax + b) dx =1aF (ax + b) , kde F je funkce primitivn k funkci f na in-

tervalu I . Plat pro ta x, pro kter je ax + b I .

Vta 5. Nech funkce f m derivaci a nem nulov bod na intervalu I . Potom

na tomto intervalu plat

f(x)

f (x)dx = ln |f (x)| .

Zkladn vzorce cRobert Mak, 2012

Najdte

(2x + 3 4x +

6

x3 sinx + ex) dx.

I =

(2x + 3 4x +

6

x3 sinx + ex) dx

= 2

x dx + 3

x14 dx + 6

x3 dx

sinx dx +

ex dx

= 2x2

2+ 3

x5/4

5/4+ 6

x2

2 ( cosx) + ex + C

= x2 +125x5/4 3 1

x2+ cosx + ex + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?2*x+3*(x)^(1/4)+6/(x^3)-sin(x)+e^x;lang=cs

Najdte

(2x + 3 4x +

6

x3 sinx + ex) dx.

I =

(2x + 3 4x +

6

x3 sinx + ex) dx

= 2

x dx + 3

x14 dx + 6

x3 dx

sinx dx +

ex dx

= 2x2

2+ 3

x5/4

5/4+ 6

x2

2 ( cosx) + ex + C

= x2 +125x5/4 3 1

x2+ cosx + ex + C

Integrl ze soutu je souet integrl.

Integrl nsobku funkce je nsobek integrlu.

Nkter funkce je mono pepsat na mocninn funkce.



Najdte

(2x + 3 4x +

6

x3 sinx + ex) dx.

I =

(2x + 3 4x +

6

x3 sinx + ex) dx

= 2

x dx + 3

x14 dx + 6

x3 dx

sinx dx +

ex dx

= 2x2

2+ 3

x5/4

5/4+ 6

x2

2 ( cosx) + ex + C

= x2 +125x5/4 3 1

x2+ cosx + ex + C

xn dx =xn+1

n + 1

sinx dx = cos x

ex dx = ex



Najdte

(2x + 3 4x +

6

x3 sinx + ex) dx.

I =

(2x + 3 4x +

6

x3 sinx + ex) dx

= 2

x dx + 3

x14 dx + 6

x3 dx

sinx dx +

ex dx

= 2x2

2+ 3

x5/4

5/4+ 6

x2

2 ( cosx) + ex + C

= x2 +125x5/4 3 1

x2+ cosx + ex + C



Najdte

tgx dx.

I =

tgx dx

=

sinxcosx

dx

= sin x

cosxdx

=

(cosx)

cosxdx

= ln | cosx| + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?tan(x);lang=cs

Najdte

tgx dx.

I =

tgx dx

=

sinxcosx

dx

= sin x

cosxdx

=

(cosx)

cosxdx

= ln | cosx| + C

Pouijeme definici funkce tangens. Zkladn vzorce cRobert Mak, 2012


Najdte

tgx dx.

I =

tgx dx

=

sinxcosx

dx

= sin x

cosxdx

=

(cosx)

cosxdx

= ln | cosx| + C

Plat (cosx) = sinx. itatel se tedy li od derivace jmenovatele jenomkonstantm nsobkem.

Vynsobme a vydlme integrl tmto nsobkem.



Najdte

tgx dx.

I =

tgx dx

=

sinxcosx

dx

= sin x

cosxdx

=

(cosx)

cosxdx

= ln | cosx| + C

Formln pouijeme vztah (cosx) = sinx, abychom vidli vzorec

f(x)

f (x)dx = ln |f (x)|+ C.



Najdte

tgx dx.

I =

tgx dx

=

sinxcosx

dx

= sin x

cosxdx

=

(cosx)

cosxdx

= ln | cosx| + C

f(x)

f (x)dx = ln |f (x)|+ C



Najdte

x + 2

x2 + 4x + 5dx.

I =

x + 2

x2 + 4x + 5dx

=12

2x + 4

x2 + 4x + 5dx

=12

(x2 + 4x + 5)

x2 + 4x + 5dx

=12

ln(x2 + 4x + 5) + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+2)/(x^2+4*x+5);lang=cs

Najdte

x + 2

x2 + 4x + 5dx.

I =

x + 2

x2 + 4x + 5dx

=12

2x + 4

x2 + 4x + 5dx

=12

(x2 + 4x + 5)

x2 + 4x + 5dx

=12

ln(x2 + 4x + 5) + C

Plat (x2 + 4x + 5) = 2x + 4. itatel se tedy li od derivace jmenovatelejenom konstantm nsobkem.

Vynsobme a vydlme integrl tmto nsobkem.



Najdte

x + 2

x2 + 4x + 5dx.

I =

x + 2

x2 + 4x + 5dx

=12

2x + 4

x2 + 4x + 5dx

=12

(x2 + 4x + 5)

x2 + 4x + 5dx

=12

ln(x2 + 4x + 5) + C

Pepeme do tvaru

f(x)

f (x)dx.



Najdte

x + 2

x2 + 4x + 5dx.

I =

x + 2

x2 + 4x + 5dx

=12

2x + 4

x2 + 4x + 5dx

=12

(x2 + 4x + 5)

x2 + 4x + 5dx

=12

ln(x2 + 4x + 5) + C

f(x)

f (x)dx = ln |f (x)|+ C



Najdte

x + 5

x2 + 4dx.

I =

x + 5

x2 + 4dx

=

12 2xx2 + 4

+5

x2 + 4dx

=12

ln(x2 + 4) + 512

arctgx

2+ C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2+4);lang=cs

Najdte

x + 5

x2 + 4dx.

I =

x + 5

x2 + 4dx

=

12 2xx2 + 4

+5

x2 + 4dx

=12

ln(x2 + 4) + 512

arctgx

2+ C

Derivace jmenovatele je x, v itateli vak nen nsobek tto funkce.

Vzorec

f(x)

f (x)dx nelze pmo pout.

Rozdlme zlomek na dva. Zkladn vzorce cRobert Mak, 2012


Najdte

x + 5

x2 + 4dx.

I =

x + 5

x2 + 4dx

=

12 2xx2 + 4

+5

x2 + 4dx

=12

ln(x2 + 4) + 512

arctgx

2+ C

V prvnm zlomku je v itateli polovina derivace jmenovatele.

Proto prvn zlomek vynsobme a vydlme dvma.



Najdte

x + 5

x2 + 4dx.

I =

x + 5

x2 + 4dx

=

12 2xx2 + 4

+5

x2 + 4dx

=12

ln(x2 + 4) + 512

arctgx

2+ C

f(x)

f (x)= ln |f (x)|+ C

1

A2 + x2dx =

1A

arctgx

A



Najdte

1

(x + 6)3dx.

I =

1

(x + 6)3dx

=

(x + 6)3 dx

=(x + 6)2

2= 1

2(x + 6)2+ C


http://wood.mendelu.cz/math/maw/integral/integralx.php?1/(x+6)^3;lang=cs

Najdte

1

(x + 6)3dx.

I =

1

(x + 6)3dx

=

(x + 6)3 dx

=(x + 6)2

2= 1

2(x + 6)2+ C

Jedn se o mocninnou funkci. Zkladn vzorce cRobert Mak, 2012


Najdte

1

(x + 6)3dx.

I =

1

(x + 6)3dx

=

(x + 6)3 dx

=(x + 6)2

2= 1

2(x + 6)2+ C

f (ax + b) dx =1aF (ax + b), kde F je integrl z f .

V naem ppad je f (x) = x3, F (x) = x2

2 a a = 1.



Najdte

1

(x + 6)3dx.

I =

1

(x + 6)3dx

=

(x + 6)3 dx

=(x + 6)2

2= 1

2(x + 6)2+ C

Upravme. Zkladn vzorce cRobert Mak, 2012


Najdte nsledujc integrly.

12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C



12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C

1x

dx = ln |x|

f (ax + b) dx =1aF (ax + b), v naem ppad a = 2.



12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C

Pepeme na mocninnou funkci. Zkladn vzorce cRobert Mak, 2012


12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C

xn dx =1

n + 1xn+1




12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C



12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C

ex dx = ex




12x + 5

dx =12

ln |2x + 5| + C

1

(2 x)5dx =

(2 1 x)5 dx

=(2 x)4

4 11

=1

4(2 x)4+ C

ex dx = ex + C

e3x dx =13e3x + C

ex dx = ex




(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

Upravme podle vzorce (a + b)2:

(ex + ex)2 = e2x + 2exex + e2x = e2x + 2 + e2x



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

Integrujeme podle vzorc

ex dx = ex ,

1 dx = x ,

f (ax + b) dx =1aF (ax + b) , kde

f (x) dx = F (x).



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

Pouijeme vzorecsin(2x) = 2 sinx cosx



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

Integrujeme podle vzorc

sinx dx = cosx

a

f (ax + b) dx =1aF (ax + b) , kde

f (x) dx = F (x).



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

Vzorec

sin2 x =1 cos(2x)

2



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

cosx dx = sinx

f (ax + b) =1aF (ax + b)



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

Potebujeme vydlit. K tomu je mono pevst itatel na tvar, kter pozdjiumon zkrtit. Zkladn vzorce cRobert Mak, 2012


(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

x2 1 + 1x + 1

=x2 1x + 1

+1

x + 1= x 1 + 1

x + 1



(ex + ex)2 dx =

(e2x + 2 + e2x) dx

=12e2x + 2x1

2e2x + C

sinx cosx dx =12

sin(2x) dx =12 12 ( cos 2x) + C

sin2 x dx =12

(

1 cos(2x))

dx =12

[

x 12

sin(2x)]

+ C

x2

x + 1dx =

x2 1 + 1x + 1

dx =

x 1 + 1x + 1

dx

=x2

2 x + ln |x + 1| + C

xn dx =1

n + 1xn+1,

1x

dx = ln |x|,

f (ax + b) dx =1af (ax + b)


Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+5)/(x^2-4*x+9);lang=cs

Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

Zaifrujeme derivaci jmenovatele, tj. vraz (2x 4), do itatele. Zkladn vzorce cRobert Mak, 2012


Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

Musme upravit zlomek tak, aby se zlomky v prvnm a druhm integrlurovnaly.

K tmto pravm pouijeme jenom multiplikativn a aditivn konstanty(nenadlaj moc velkou neplechu pi integraci).

Pidnm nsobku 12

mme ve druhm zlomku v itateli vraz

12

(2x 4) = x 2. Koeficient u x je v podku.



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

12

(2x 4) = x 2

12

(2x 4) + 2 = x

Nyn je v itateli jenom x. Chyb slo 5.



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

12

(2x 4) = x 2

12

(2x 4) + 2 = x

12

(2x 4) + 2 + 5 = x + 5

Prvn a druh zlomek jsou stejn, nedopustili jsme se dn pravy,kter by zmnila hodnotu zlomku.



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

Rozdlme zlomek na dva. Zkladn vzorce cRobert Mak, 2012


Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

f(x)

f (x)= ln |f (x)|+ C



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

Doplnme na tverec ve jmenovateli druhho zlomku.

x2 4x + 9 = x2 2 2 x + 22 4 + 9 = (x 2)2 + 5



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

1

A2 + x2dx =

1A

arctgx

A, kde v naem ppad A =

5



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

f (ax + b) dx =1aF (ax + b), v naem ppad a = 1



Najdte

x + 5

x2 4x + 9dx.

I =

x + 5

x2 4x + 9dx

=

12 (2x 4)+2+5x2 4x + 9

dx

=

12 2x 4x2 4x + 9

+2 + 5

x2 4x + 9dx

=12

ln |x2 4x + 9| +

7

(x 2)2 + 5dx

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5 11

=12

ln |x2 4x + 9| + 7 15

arctgx 2

5+ C

Upravme. Zkladn vzorce cRobert Mak, 2012


3 Integrace per-parts

Vta 6. Nech funkce u a v maj derivace na intervalu I . Pak plat

u(x)v (x) dx = u(x)v (x)

u(x)v (x) dx, (2)

pokud integrl na prav stran existuje.

Integrace per-parts cRobert Mak, 2012

Vta 6. Nech funkce u a v maj derivace na intervalu I . Pak plat

u(x)v (x) dx = u(x)v (x)

u(x)v (x) dx, (3)

pokud integrl na prav stran existuje.Dkaz:

(uv) = uv + uv derivace souinu

(uv) dx =

uv dx +

uv dx zintegrovn a linearita integrlu

uv =

uv dx +

uv dx integrl odstran derivaci

uv

uv dx =

uv dx algebraick prava

Integrly typick pro vpo et metodou per-parts. P (x) je polynom.

P (x)ex dx,

P (x) sin(x) dx,

P (x) cos(x) dx,

P (x)arctgx dx,

P (x)lnm x dx.


Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C

Funkce je souinem polynomu a logaritmick funkce per-parts. Integrace per-parts cRobert Mak, 2012

http://wood.mendelu.cz/math/maw/integral/integralx.php?(x+1)*ln(x);lang=cs

Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C

Integrujeme per-parts pomoc vzorce

u v dx = u v

u v dx

pi u = ln x a v = x + 1. Integrace per-parts cRobert Mak, 2012


Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C


u v dx = u v

u v dx



Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C


u v dx = u v

u v dx



Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C

Roznsobme zvorku. Integrace per-parts cRobert Mak, 2012


Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C

Dokonme integraci. Integrace per-parts cRobert Mak, 2012


Vypotte

(x + 1) lnx dx

(x + 1) lnx dx =

u = ln x u =1x

v = x + 1 v =x

2

2+ x

= ln x

(

x2

2+ x

)

1x

(

x2

2+ x

)

dx

=

(

x2

2+ x

)

ln x (

12x + 1

)

dx

=

(

x2

2+ x

)

ln x (

12 x

2

2+ x

)

=

(

x2

2+ x

)

ln x 14x2 x + C



Vypotte

x sinx dx

x sinx dxu = x u = 1

v = sinx v = cosx

= x cosx

1 ( cosx) dx

= x cosx +

cosx dx

= x cosx + sinx +C


http://wood.mendelu.cz/math/maw/integral/integralx.php?x*sin(x);lang=cs

Vypotte

x sinx dx


v = sinx v = cosx

= x cosx

1 ( cosx) dx

= x cosx +

cosx dx

= x cosx + sinx +C


u v dx = u v

u v dx

pi u = x a v = sinx. Integrace per-parts cRobert Mak, 2012


Vypotte

x sinx dx


v = sinx v = cosx

= x cosx

1 ( cosx) dx

= x cosx +

cosx dx

= x cosx + sinx +C


u v dx = u v

u v dx



Vypotte

x sinx dx


v = sinx v = cosx

= x cosx

1 ( cosx) dx

= x cosx +

cosx dx

= x cosx + sinx +C


u v dx = u v

u v dx



Vypotte

x sinx dx


v = sinx v = cosx

= x cosx

1 ( cosx) dx

= x cosx +

cosx dx

= x cosx + sinx +C

Upravme. Integrace per-parts cRobert Mak, 2012


Vypotte

x sinx dx


v = sinx v = cosx

= x cosx

1 ( cosx) dx

= x cosx +

cosx dx

= x cosx + sinx +C

Integruje druhou st:

cosx dx = sinx

Hotovo. Integrace per-parts cRobert Mak, 2012


Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + C

Funkce je souinem polynomu a sinu per-parts. Integrace per-parts cRobert Mak, 2012

http://wood.mendelu.cz/math/maw/integral/integralx.php?(x-2)*sin(2*x);lang=cs

Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + CIntegrujeme per-parts pomoc vzorce

u v dx = u v

u v dx

pi u = x 2 a v = sin(2x). Integrace per-parts cRobert Mak, 2012


Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + C

Plat

v =

v (x) dx =

sin(2x) dx = 12

cos(2x),

protoe

sinx dx = cosx a

f (ax + b) =1aF (ax + b).



Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + C

u v dx = u v

u v dx



Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + C

Vytkneme konstantu(

12

)

z integrlu.



Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + CPlat

cos(2x) dx =12

sin(2x), protoe

cosx dx = sinx a

f (ax + b) =1aF (ax + b).



Vypotte

(x 2) sin(2x) dx

(x 2)sin(2x) dx =u = x 2 u = 1

v = sin(2x) v = 12

cos 2x

= (x 2) (

12

cos(2x))

1 (

12

cos 2x)

dx

= 12

(x 2) cos(2x) + 12

cos 2x dx

= 12

(x 2) cos(2x) + 12 12

sin(2x) + C

= 12

(x 2) cos(2x) + 14

sin(2x) + C



Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))

= (1 x2) cosx + 2x sinx + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*sin(x);lang=cs

Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))


Funkce je souinem polynomu a funkce sinus.

Budeme integrovat per-parts podle vzorce

u v dx = u v

u v dx

pi volb u = (x2 + 1) a v = sinx.



Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))

= (1 x2) cosx + 2x sinx + C(x2 + 1) = 2x

sinx dx = cosx



Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))


u v dx = u v

u v dx

Konstantn nsobek 2 a znamnko minus dme ped integrl. Integrace per-parts cRobert Mak, 2012


Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))


Jet jednou integrujeme per-parts. Nyn u = x a v = cosx. Integrace per-parts cRobert Mak, 2012


Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))

= (1 x2) cosx + 2x sinx + Cx = 1

cosx dx = sinx



Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))

= (1 x2) cosx + 2x sinx + Cu v dx = u v

u v dx



Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))

= (1 x2) cosx + 2x sinx + CIntegrujeme sinus:

sinx dx = cosx Integrace per-parts cRobert Mak, 2012


Vypotte

(x2 + 1) sinx dx

(x2 + 1) sinx dxu = x2 + 1 u = 2x

v = sinx v = cosx

= (x2 + 1) cosx + 2

x cosx dx

u = x u = 1

v = cosx v = sinx

= (x2 + 1) cosx + 2(

x sinx

sinx dx)

= (x2 + 1) cosx + 2(

x sinx ( cosx))




Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,


http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^2+1)*exp(-x);lang=cs

Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,

Integruje souin polynomu a exponenciln funkce. Integrace per-parts cRobert Mak, 2012


Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,

Integrujeme per-parts.

Polynom budeme derivovat a exponencielu integrovat.

Nezapomeme, e

ex dx = ex.



Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,Vzorec je

u v dx = u v

u v dx. Integrace per-parts cRobert Mak, 2012


Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C, Opt polynom krt exponenciln funkce.

Opt integrujeme per-parts. Opt derivujeme polynom.



Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,

Vzorec pro ervenou st je

uv dx = uv

uv dx, zbytek zstane.



Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,

ex dx = ex



Vypotte

(x2 + 1)ex dx.

(x2 + 1)ex dxu = x2 + 1 u = 2x

v = ex v = ex

= (x2 + 1)ex + 2

xex dxu = x u = 1

v = ex v = ex

= (x2 + 1)ex + 2(

xex +

ex dx)

= (x2 + 1)ex + 2(xex ex) = ex(x2 + 2x + 3) + C,

Vytkneme (ex). Integrace per-parts cRobert Mak, 2012


Vypotte

x arctgx dx.

x arctgx dx

u = arctgx u =1

1 + x2

v = x v =x

2

2

=x2

2arctgx 1

2

x2

1 + x2dx

=x2

2arctgx 1

2

1 11 + x2

dx

=x2

2arctgx 1

2

(

x arctgx)

+ C.


http://wood.mendelu.cz/math/maw/integral/integralx.php?x*atan(x);lang=cs

Vypotte

x arctgx dx.

x arctgx dx

u = arctgx u =1

1 + x2

v = x v =x

2

2

=x2

2arctgx 1

2

x2

1 + x2dx

=x2

2arctgx 1

2

1 11 + x2

dx

=x2

2arctgx 1

2

(

x arctgx)

+ C.

Jedn se o souin polynomu a funkce arkustangens. Integrace per-parts cRobert Mak, 2012


Vypotte

x arctgx dx.

x arctgx dx

u = arctgx u =1

1 + x2

v = x v =x

2

2

=x2

2arctgx 1

2

x2

1 + x2dx

=x2

2arctgx 1

2

1 11 + x2

dx

=x2

2arctgx 1

2

(

x arctgx)

+ C.

Budeme integrovat metodou per-parts. Budeme integrovat polynom aderivovat arkustangens. Integrace per-parts cRobert Mak, 2012


Vypotte

x arctgx dx.

x arctgx dx

u = arctgx u =1

1 + x2

v = x v =x

2

2

=x2

2arctgx 1

2

x2

1 + x2dx

=x2

2arctgx 1

2

1 11 + x2

dx

=x2

2arctgx 1

2

(

x arctgx)

+ C.

uv dx = uv

uv dx



Vypotte

x arctgx dx.

x arctgx dx

u = arctgx u =1

1 + x2

v = x v =x

2

2

=x2

2arctgx 1

2

x2

1 + x2dx

=x2

2arctgx 1

2

1 11 + x2

dx

=x2

2arctgx 1

2

(

x arctgx)

+ C.

Musme integrovat racionln funkci. Nejprve provedeme dlen:

x2

x2 + 1=

(x2 + 1) 1x2 + 1

=x2 + 1

x2 + 1 1

x2 + 1= 1 1

x2 + 1



Vypotte

x arctgx dx.

x arctgx dx

u = arctgx u =1

1 + x2

v = x v =x

2

2

=x2

2arctgx 1

2

x2

1 + x2dx

=x2

2arctgx 1

2

1 11 + x2

dx

=x2

2arctgx 1

2

(

x arctgx)

+ C.

K dokonen zbv integrovat jedniku a jeden zlomek. To provedemepomoc pslunch vzorc. Integrace per-parts cRobert Mak, 2012


Vypotte

ln x dx

1 ln x dxu = lnx u =

1x

v = 1 v = x

= x lnx

1 dx

= x lnx x= x(lnx 1) + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?ln(x);lang=cs

Vypotte

ln x dx


1x

v = 1 v = x

= x lnx

1 dx


Ve funkci je zaifrovan souin polynomu a logaritmick funkce:

1 ln x dx.

Integrujeme per-parts pi volb u = lnx a v = 1. Integrace per-parts cRobert Mak, 2012


Vypotte

ln x dx


1x

v = 1 v = x

= x lnx

1 dx


(lnx) =1x

1 dx = x



Vypotte

ln x dx


1x

v = 1 v = x

= x lnx

1 dx


u v dx = u v

u v dx

Uijeme vztah1xx = 1.



Vypotte

ln x dx


1x

v = 1 v = x

= x lnx

1 dx


1 dx = x



Vypotte

ln x dx


1x

v = 1 v = x

= x lnx

1 dx


Hotovo. Integrace per-parts cRobert Mak, 2012


Vypotte

ln2 x dx

1 ln2 x dxu = ln2 x u =

2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)

= x ln2 x 2x lnx + 2x + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(ln(x))^2;lang=cs

Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)

= x ln2 x 2x lnx + 2x + C Je zde zaifrovn souin polynomu a druh mocniny logaritmu.

Upravme funkci ln2 x na souin (1) (ln2 x) a integrujeme per-parts pivolb u = ln2 x a v = 1



Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)

= x ln2 x 2x lnx + 2x + C(ln2 x) = 2 ln x(lnx) = 2 lnx

1x

1 dx = x



Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)


u v dx = u v

u v dx



Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)


Tento trik ji znme: Napeme funkci ln x jako souin (1) lnx a integrujemeper-parts pi volb u = lnx a v = 1. Integrace per-parts cRobert Mak, 2012


Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)

= x ln2 x 2x lnx + 2x + C(lnx) =

1x

1 dx = x



Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)


u v dx = u v

u v dx



Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)


Dopotme integrl z jedniky. Integrace per-parts cRobert Mak, 2012


Vypotte

ln2 x dx


2 ln xx

v = 1 v = x

= x ln2 x 2

ln x dx

u = ln x u =1x

v = 1 v = x

= x ln2 x 2(

x lnx

1 dx)

= x ln2 x 2(

x lnx x)


Upravme. Hotovo. Integrace per-parts cRobert Mak, 2012


Najdte

x3 sinx dx.

x3 sinx dx =

u = x3 3x2 6x 6 0

v = sinx cos x sin x cosx sinx= x3 cosx (3x2 sinx) + 6x cosx 6 sinx= (x3 + 6x) cos(x) + (3x2 6) sinx + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?x^3*sin(x);lang=cs

Najdte

x3 sinx dx.

x3 sinx dx =

u = x3 3x2 6x 6 0


Tikrt integrujeme per-parts, ale vechno zapeme do jednoho sche-matu.

lut ipka reprezentuje derivovn. Derivujeme a na nulu.

erven ipka reprezentuje integrovn.

derivace derivace derivace derivace

integrace integrace integrace integrace



Najdte

x3 sinx dx.

x3 sinx dx =

u = x3 3x2 6x 6 0


Nsobme ve smru ipek. Souinm ve smru lutch ipek znamnkoponechme, souinm ve smru ervench ipek znamnko zmnme avechny souiny seteme.

souin

souinsouin

souin



Najdte

x3 sinx dx.

x3 sinx dx =

u = x3 3x2 6x 6 0




Najdte

(x3 + 2x)ex dx.

(x3 + 2x)ex dx

=

u = x3+ 2x 3x2 + 2 6x 6 0

v = ex ex ex ex ex

= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex

= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8)


http://wood.mendelu.cz/math/maw/integral/integralx.php?(x^3+2x)*exp(-x);lang=cs

Najdte

(x3 + 2x)ex dx.

(x3 + 2x)ex dx

=

u = x3+ 2x 3x2 + 2 6x 6 0

v = ex ex ex ex ex

= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex

= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8) Tikrt integrujeme per-parts, ale vechno zapeme do jednoho sche-matu.

lut ipka reprezentuje derivovn. Derivujeme a na nulu.

erven ipka reprezentuje integrovn.

derivace derivace derivace derivace

integrace integrace integrace integrace



Najdte

(x3 + 2x)ex dx.

(x3 + 2x)ex dx

=

u = x3+ 2x 3x2 + 2 6x 6 0

v = ex ex ex ex ex

= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex

= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8)

Nsobme ve smru ipek. Souinm ve smru lutch ipek znamnkoponechme, souinm ve smru ervench ipek znamnko zmnme avechny souiny seteme.

souin

souinsouin

souin



Najdte

(x3 + 2x)ex dx.

(x3 + 2x)ex dx

=

u = x3+ 2x 3x2 + 2 6x 6 0

v = ex ex ex ex ex

= (x3 + 2x)ex (3x2 + 2)ex + (6xex) 6ex

= ex(x3 + 2x + 3x2 + 2 + 6x + 6)= ex(x3 + 3x2 + 8x + 8)



4 Integrace pomoc substituce.

Vta 7. Nech f (t) je funkce spojit na intervalu I , nech funkce (x) m deri-vaci na intervalu J a plat (J) = I . Potom na intervalu J plat

f ((x))(x) dx =

f (t) dt, (4)

dosadme-li napravo t = (x).

Schematicky: (x) = t (x) dx = dt

Vta 8. Nech f (x) je funkce spojit na intervalu I , nech funkce (t) m nenu-lovou derivaci na intervalu J a plat (J) = I . Potom na intervalu I plat

f (x) dx =

f ((t))(t) dt, (5)

dosadme-li napravo t = 1(x), kde 1(x) je funkce inverzn k funkci (x).

Schematicky: x = (t) dx = (t) dt

Integrace pomoc substituce. cRobert Mak, 2012

Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt

= cos t = cos(lnx) + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?sin(ln(x))/x;lang=cs

Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt


Vnitn sloka je ln x.

Derivace funkce ln x je 1x

.

Tato derivace, 1x

, je v souinu s integrovanou funkc.



Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt


Zavedeme substituci lnx = t. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt


Nalezneme vztah mezi dx a dt. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt


Dosadme substituci. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt


Integrujeme. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

sin(ln x)x

dx

sin(lnx)x

dx =

sin(lnx)1x

dxlnx = t

1x

dx = dt=

sin t dt


Pouijeme substituci k nvratu k promnn x a pidme integran konstantu.Hotovo. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2


http://wood.mendelu.cz/math/maw/integral/integralx.php?x*exp(1-x^2);lang=cs

Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Zkusme substituovat za vnitn sloku sloen funkce e1x2

. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Hledme vztah mezi diferencily. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Derivujeme ob strany substituce. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Vyjdme odsud vraz x dx, kter figuruje uvnit integrlu. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Dosadme. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Vypotte integrl pomoc vzorce. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

xe1x2

dx.

xe1x2

dx

1 x2 = t2x dx = dtx dx = 1

2dt

= 12

et dt

= 12et

= 12e1x

2

Pouijeme substituci pro nvrat k pvodn promnn. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C


http://wood.mendelu.cz/math/maw/integral/integralx.php?x/(x^4+16);lang=cs

Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C

Substituce x4+16 = t, nebo x4 = t, nejsou pln ikovn, protoe vztahmezi diferencily pi tto substituci je

4x3 dx = dt,

avak len x3 dx nikde v integrlu nen.

len x dx napovd, pout substituci x2 = t.



Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C

Hledme vztah mezi diferencily a vyjdme z nj vraz x dx. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C

Substituce x2 = t vede k relaci x4 = (x2)2 = t2. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C



Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C

Uijeme vzorec

1

x2 + A2dx =

1A

arctgx

A

pi A = 4. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

x

x4 + 16dx

x

x4 + 16dx

x2 = t

2x dx = dt

x dx =12

dt

x4 = t2

=12

1

t2 + 16dt

=18

arctgt

4=

18

arctgx

2

4+ C

Uijeme zptnou substituci t = x2. Hotovo. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?exp(sqrt(x+1))/sqrt(x+1);lang=cs

Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C

Vnitn sloka je

x + 1. Derivace tto vnitn sloky je

(

x + 1) =12

(x + 1)1/2 =12 1

x + 1.

Vskyt tto lene1

x + 1

uvnit integrlu (a v souinu) napovd, e provst

tuto substituci bude snadn. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C

Pouijeme navrenou substituci. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C

Najdeme vztah mezi diferencily dx a dt. Dostvme

12

1x + 1

dx = dt

a tuto relaci vynsobme slem 2. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C



Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C

Zintegrujeme. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

ex+1

x + 1

dx

ex+1

x + 1

dx =

ex+1 1

x + 1dx

x + 1 = t1

2x + 1

dx = dt

1x + 1

dx = 2 dt

=

et2 dt = 2et = 2ex+1 + C

Uijeme substituci t =

x + 1 k nvratu k pvodn promnn. Hotovo. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t

sinx dx = dtsinx dx = dt

=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(tan(x))^3;lang=cs

Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Rozepeme funkci tgx pomoc funkc sinx a cosx.

Lich mocnina je i v itateli, i ve jmenovateli. Vybereme si tu v itateli.



Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Vythneme jednu mocninu funkce sinx z itatele. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Sudou mocninu pevedeme na funkci cosx. Uijeme identitusin2 x + cos2 x = 1. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Dosadme cosx = t. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Nalezneme vztah mezi diferencily dx a dt. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Pepeme vraz sinx dx do novch promnnch. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C



Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Upravme Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Protoe je jmenovatel jednolenn, sta vydlit itatele vrazem t3. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Nyn integrujeme pomoc vzorc. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

tg3 x dx.

tg3 x dx =

sin3 x

cos3 xdx =

sin2 x

cos3 xsinx dx =

1 cos2xcos3x

sinx dx

cosx = t


=

1 t2

t3dt =

t2 1t3

dt =

1t t3 dt

= ln |t| + 12t2 = ln | cosx| + 1

2 cos2 x+ C

Po integraci provedeme nvrat k pvodn promnn a pidme integrankonstantu. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12

ln |t2 + 1| arctg t)

+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C


http://wood.mendelu.cz/math/maw/integral/integralx.php?(sqrt(3*x+2)-1)/(x+1);lang=cs

Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C len 3x + 2 je pod odmocninou. Uijeme substituci, kter umon tuto

odmocninu odstranit.

Budeme dosazovat 3x + 2 = t2. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Nalezneme vztah mezi dx a dt. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Vyjdme dx. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Vyjdme promnnou x. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Pichystme si zptnou substituci. Vyjdme t pomoc x. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Provedeme substituci. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Upravme. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Pevedeme na jeden zlomek. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + CVydlmet

2 tt2 + 1

=(t2 + 1) + (t 1)

t2 + 1= 1 +

t 1t2 + 1



Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + CZskan funkce je zlomek, kter ped integrovnm rozdlme na souetzlomku, kter m v itateli derivaci jmenovatele, a zlomku, kter m v itatelijen konstantu. Oba zlomky pak snadno zintegrujeme. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + CIntegrace je ji snadn. Uijeme vztah

t

t2 + 1dt =

12

2t

t2 + 1dt =

12

ln(t2 + 1).



Vypotte

3x + 2 1x + 1

dx.

3x + 2 1x + 1

dx

3x + 2 = t2

3 dx = 2t dt

dx =23t dt

x =13

(t2 2)

t =

3x + 2

=

t 113 (t

2 2) + 1 23t dt

= 2

t 1t2 + 1

t dt = 2

t2 tt2 + 1

dt = 2

1 +t 1t2 + 1

dt

= 2

1 tt2 + 1

1t2 + 1

dt = 2(

t 12


+ C

= 2

3x + 2 ln |3x + 3| 2 arctg

3x + 2 + C

Roznsobme zvorku a provedeme zptnou substituci pro nvrat kpromnn x. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

1 +x 1x

dx.

1 +x 1x

dx =

x 1 = t2

dx = 2t dt

x = t2 + 1

x 1 = t

=

1 +t2

t2 + 1 2t dt

= 2

1 + t

t2 + 1 t dt = 2

t2 + t

t2 + 1dt = 2

1 +t 1t2 + 1

dt

= 2

1 +12 2tt2 + 1

1t2 + 1

dt

= 2[

t +12

ln |t2 + 1| arctg t]

= 2[

x 1 + 12

ln |x| arctg

x 1]

+ C

Funkce obsahuje odmocninu z linernho vrazu zavedeme substituci naodstrann odmocniny. Integrace pomoc substituce. cRobert Mak, 2012

http://wood.mendelu.cz/math/maw/integral/integralx.php?(1+sqrt(x-1))/x;lang=cs

Vypotte

1 +x 1x

dx.

1 +x 1x

dx =

x 1 = t2

dx = 2t dt

x = t2 + 1

x 1 = t

=

1 +t2

t2 + 1 2t dt

= 2

1 + t

t2 + 1 t dt = 2

t2 + t

t2 + 1dt = 2

1 +t 1t2 + 1

dt

= 2

1 +12 2tt2 + 1

1t2 + 1

dt

= 2[

t +12


= 2[

x 1 + 12

ln |x| arctg

x 1]

+ C

Vraz pod odmocninou je druh mocnina nov promnn. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

1 +x 1x

dx.

1 +x 1x

dx =

x 1 = t2

dx = 2t dt

x = t2 + 1

x 1 = t

=

1 +t2

t2 + 1 2t dt

= 2

1 + t

t2 + 1 t dt = 2

t2 + t

t2 + 1dt = 2

1 +t 1t2 + 1

dt

= 2

1 +12 2tt2 + 1

1t2 + 1

dt

= 2[

t +12


= 2[

x 1 + 12

ln |x| arctg

x 1]

+ CNalezneme vztah mezi diferencily dx a dt

(x 1) = 1 (derivace podle x)

(t2) = 2t (derivace podle t) Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

1 +x 1x

dx.

1 +x 1x

dx =

x 1 = t2

dx = 2t dt

x = t2 + 1

x 1 = t

=

1 +t2

t2 + 1 2t dt

= 2

1 + t

t2 + 1 t dt = 2

t2 + t

t2 + 1dt = 2

1 +t 1t2 + 1

dt

= 2

1 +12 2tt2 + 1

1t2 + 1

dt

= 2[

t +12


= 2[

x 1 + 12

ln |x| arctg

x 1]

+ C

Nalezneme x a

x 1 ze substitunho vztahu. Integrace pomoc substituce. cRobert Mak, 2012


Vypotte

1 +x 1x

dx.

1 +x 1x

dx =

x 1 = t2

dx

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