# Integration Techniques' Examples

## Antidifferentiation

Example 1 :
Question:

Find the antiderivative for the following function.
f(x) = ${\displaystyle e^{3x}}$
Solution:
∫ f(x)dx
${\displaystyle e^{3x}}$dx
=${\displaystyle {\frac {1}{3}}e^{3x}}$
=${\displaystyle {\frac {e^{3x}}{3}}}$ + C

Example 2 :
Question:

Find the antiderivative for the following function.
f(x) = -4x
Solution:
∫f(x)dx
∫ -4x dx
=${\displaystyle -2x^{2}}$ + C

Example 3 :
Question:

∫4${\displaystyle x^{3}}$dx
Solution:
= ${\displaystyle x^{4}}$ + C

## Integration By Substitution

Example 1 :
Question:

Determine:
${\displaystyle {(x^{2}+1)}^{3}2xdx}$
Solution:
Set:
${\displaystyle u=x^{2}+1}$
${\displaystyle du={\frac {d}{dx}}(x^{2}+1)dx}$
${\displaystyle du=2xdx}$
${\displaystyle {(x^{2}+1)}^{3}2xdx}$
${\displaystyle =u^{3}du}$
= ${\displaystyle {\frac {1}{4}}u^{4}+C}$
= ${\displaystyle {\frac {1}{4}}{(x^{2}+1)}^{4}+C}$

Example 2 :
Question:

Determine:
${\displaystyle {\frac {({lnx}^{2})}{x}}dx}$
Solution:
Set:
${\displaystyle u=lnx}$
${\displaystyle du={\frac {1}{x}}dx}$
${\displaystyle {\frac {({lnx}^{2})}{x}}dx}$
${\displaystyle =({lnx}^{2}){\frac {1}{x}}dx}$
= ∫${\displaystyle u^{2}du}$
= ${\displaystyle {\frac {u^{3}}{3}}+C}$
= ${\displaystyle {\frac {(lnx)^{3}}{3}}+C}$

## Integration By Parts

Example 1 :
Question:

${\displaystyle xe^{x}dx}$
Solution:
∫ f(x)g(x)dx = f(x)G(x) - ∫f'(x)G(x)dx
Set:
f(x) = x
f'(x) = 1
g(x) = ${\displaystyle e^{x}}$
G(x) = ${\displaystyle e^{x}}$
${\displaystyle xe^{x}dx}$
= ${\displaystyle xe^{x}-}$${\displaystyle 1(e^{x})dx}$
= ${\displaystyle xe^{x}-e^{x}+C}$

Example 2 :
Question:

${\displaystyle x(x+5)^{8}dx}$
Solution:
∫ f(x)g(x)dx = f(x)G(x) - ∫f'(x)G(x)dx
Set:
f(x) = x
f'(x) = 1
g(x) = ${\displaystyle (x+5)^{8}}$
G(x) = ${\displaystyle {\frac {1}{9}}(x+5)^{9}}$
${\displaystyle x(x+5)^{8}dx}$
= ${\displaystyle x{\frac {1}{9}}(x+5)^{9}-}$${\displaystyle 1{\frac {1}{9}}(x+5)^{9}dx}$
= ${\displaystyle {\frac {1}{9}}x(x+5)^{9}-{\frac {1}{9}}}$${\displaystyle (x+5)^{9}dx}$
= ${\displaystyle {\frac {1}{9}}x(x+5)^{9}-{\frac {1}{9}}{\frac {1}{10}}(x+5)^{10}+C}$
= ${\displaystyle {\frac {1}{9}}x(x+5)^{9}-{\frac {1}{90}}(x+5)^{10}+C}$

Example 3 :
Question:

${\displaystyle xsinxdx}$
Solution:
∫ f(x)g(x)dx = f(x)G(x) - ∫f'(x)G(x)dx
Set:
f(x) = x
f'(x) = 1
g(x) = sinx
G(x) = -cosx
∫xsinxdx
= -xcosx - ∫1(-cosx)dx
= -xcosx + ∫cosxdx
= -xcosx + sinx + C