basic integration

**slaypullingcat** · May 3rd 2009, 07:26 PM

Hi, hope someone can help with this. Any working or explanation would be greatly appreciated. I really want to understand why/what I need to do/think. Thank you.

Find the indefinite integral of dx/x^2 -9

Cheers

**TheEmptySet** · May 3rd 2009, 07:32 PM

Originally Posted by slaypullingcat

Hi, hope someone can help with this. Any working or explanation would be greatly appreciated. I really want to understand why/what I need to do/think. Thank you.

Find the indefinite integral of dx/x^2 -9

Cheers

I assume you mean this

$\int \frac{dx}{x^2-9}$

The first thing to do is to factor the denominator and then use partial fraction decomp

$\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}$

Clearing the fractions we get

$1=A(x+3)+B(x-3)$

if we let $x=3$ we get

$1=A(6) \iff A=\frac{1}{6}$

if we let $x=-3$ we get

$1=B(-6) \iff B=-\frac{1}{6}$

Now we have the integrals

$\int \frac{dx}{x^2-9}=\int \frac{\frac{1}{6}}{(x-3)}dx+\int \frac{-\frac{1}{6}}{(x+3)}dx$

The latter are both u subs for the denominator

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