Integration by Parts
Integration by parts is used when integrating the
product of two expressions . We can also sometimes
use integration by parts when we want to integrate a
function that cannot be split into the product of two
things.

udv = uv - ò v du
ò
Let’s see some examples

ò x cos x dx =
u=x
du = dx

dv = cos x dx
Sdv = Scos x dx
v = sin x

xsin x - ò sin x dx = xsin x - (-cos x ) + C
More Examples

ò xe

x

dx =

u=x
du = dx

xe - ò e dx =
x

x

xe x - e x + C

dv = ex dx
Sdv = Sex dx
v = ex
One More !!!

ò ln x dx =
u = ln x
du = 1/x dx

x ln x - ò dx =

dv = dx
S dv = S dx
v=x

x ln x - x +C
Repeated Integration by Parts

òxe

2 -x

u = x2
du = 2x dx

dx = x (-e
2

dv = e-x dx
Sdv =S e-x dx
v = -e-x

-x

) - ò -e (2x) dx =
-x

-x 2 e-x + 2 ò xe- x dx =

é-xe- x - ò -e- x dxù =
-x e + 2 ë
û
2 -x

u=x
du = dx

dv = e-x dx
Sdv =S e-x dx
v = -e-x

-x 2 e-x + 2 (-xe- x - e- x ) + C = -x 2 e-x - 2xe-x - 2e- x + C =

-e- x ( x 2 + 2x + 2) + C
Integration by Parts for Definite
Integrals
b

b

udv = uv] a - ò v du
ò
a

b

a
Example
1

ò tan

-1

x dx =

0

½
I 1

x tan x ù û0
1

-1

2
I

x
ò 1+ x 2 dx =
0
x2

t=1+
dt = 2x dx

ù -1
x tan xû
0
2
-1

1

ò

u = tan -1 x
1
du =
dx
2
1+ x

dv = dx
v=x

1 1 2x
x tan -1 xù - ò
dx =
û0
2
2 0 1+ x
ù
1
1
-1 ù1
dx = x tan xû0 - lnuú =
û
2
u
1

ù - 1 ln (1+ x 2 )ù = tan-1 1- 1 ( ln 2 - ln1) = p - 1 ln 2
x tan xû
ú
0
û0
2
2
4 2
-1

1

1

8.2 integration by parts

  • 1.
  • 2.
    Integration by partsis used when integrating the product of two expressions . We can also sometimes use integration by parts when we want to integrate a function that cannot be split into the product of two things. udv = uv - ò v du ò
  • 3.
    Let’s see someexamples ò x cos x dx = u=x du = dx dv = cos x dx Sdv = Scos x dx v = sin x xsin x - ò sin x dx = xsin x - (-cos x ) + C
  • 4.
    More Examples ò xe x dx= u=x du = dx xe - ò e dx = x x xe x - e x + C dv = ex dx Sdv = Sex dx v = ex
  • 5.
    One More !!! òln x dx = u = ln x du = 1/x dx x ln x - ò dx = dv = dx S dv = S dx v=x x ln x - x +C
  • 6.
    Repeated Integration byParts òxe 2 -x u = x2 du = 2x dx dx = x (-e 2 dv = e-x dx Sdv =S e-x dx v = -e-x -x ) - ò -e (2x) dx = -x -x 2 e-x + 2 ò xe- x dx = é-xe- x - ò -e- x dxù = -x e + 2 ë û 2 -x u=x du = dx dv = e-x dx Sdv =S e-x dx v = -e-x -x 2 e-x + 2 (-xe- x - e- x ) + C = -x 2 e-x - 2xe-x - 2e- x + C = -e- x ( x 2 + 2x + 2) + C
  • 7.
    Integration by Partsfor Definite Integrals b b udv = uv] a - ò v du ò a b a
  • 8.
    Example 1 ò tan -1 x dx= 0 ½ I 1 x tan x ù û0 1 -1 2 I x ò 1+ x 2 dx = 0 x2 t=1+ dt = 2x dx ù -1 x tan xû 0 2 -1 1 ò u = tan -1 x 1 du = dx 2 1+ x dv = dx v=x 1 1 2x x tan -1 xù - ò dx = û0 2 2 0 1+ x ù 1 1 -1 ù1 dx = x tan xû0 - lnuú = û 2 u 1 ù - 1 ln (1+ x 2 )ù = tan-1 1- 1 ( ln 2 - ln1) = p - 1 ln 2 x tan xû ú 0 û0 2 2 4 2 -1 1 1