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Integration by parts is simply the product rule reversed. The formula is as follows:

Formula

Integration by Parts

uvdx=uvuvdx

Note! In exercises with integration by parts, you should choose ex as v and ln(x) as u.

Example 1

3xexdx=3xex3exdx=3xex3ex+C=3ex(x1)+C

*

u=3xv=exu=3v=ex

Example 2

Find the function F such that F(x)=4x3+1x and F(1)=2

F(x)=4x3+1xdx=x4+ln|x|+C

Furthermore, given that F(1)=2:

2=F(1)=14+ln|1|+C=1+0+C,C=1

Then, F(x)=x4+ln|x|+1.

Example 3

Compute the integral sin2xdx

sin2xdx=sinxsinxdx=sinxcosx=+cos2xdx=sinxcosx=+1sin2xdx=sinxcosx=+xsin2xdx

sin2xdx=sinxsinxdx=sinxcosx+cos2xdx=sinxcosx+1sin2xdx=sinxcosx+xsin2xdx

*

u=sinxv=sinxu=cosxv=cosx

This gives you an equation that you solve for sin2xdx:

sin2xdx=sinxcosx=+xsin2xdx

2sin2xdx2=sinxcosx+x2sin2xdx=12sinxcosx+x2+C

sin2xdx=sinxcosx+xsin2xdx2sin2xdx=sinxcosx+x|:2sin2xdx=12sinxcosx+x2+C

Example 4

Compute cos(2x)sin(2x)dx

=cos(2x)sin(2x)dx=12cos2(2x)sin(2x)cos(2x)dx

cos(2x)sin(2x)dx=12cos2(2x)sin(2x)cos(2x)dx

*

u=cos(2x)v=sin(2x)u=2sin(2x)v=12cos(2x)

You now solve this expression as an equation with respect to cos(2x)sin(2x)dx:

cos(2x)sin(2x)dx=12cos2(2x)sin(2x)cos(2x)dx2cos(2x)sin(2x)dx=12cos2(2x)|÷2cos(2x)sin(2x)dx=14cos2(2x)+C

cos(2x)sin(2x)dx=12cos2(2x)sin(2x)cos(2x)dx2cos(2x)sin(2x)dx=12cos2(2x)|÷2cos(2x)sin(2x)dx=14cos2(2x)+C

Example 5

Compute ex(x2+3x4)dx

=ex(x2+3x4)dx=ex(x2+3x4)ex(2x+3)dx=ex(x2+3x4)(ex(2x+3)2exdx)=ex(x2+3x4)ex(2x+3)+2ex+C=ex(x2+x5)+C

ex(x2+3x4)dx=ex(x2+3x4)ex(2x+3)dx=ex(x2+3x4)(ex(2x+3)2exdx)=ex(x2+3x4)ex(2x+3)+2ex+C=ex(x2+x5)+C

*

u=x2+3x4v=exu=2x+3v=ex

**

z=2x+3w=exz=2w=ex

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