Question

How do you integrate `1/(x^2-4)` using partial fractions?

Answer 1

The integral is `=1/4(ln(x-2)-ln(x+2))+C`

Explanation:

Let factorise the denominator
`x^2-4=(x-2)(x+2)`
So we have
`1/(x^2-4)=1/((x-2)(x+2))`
So the decomposition into partial fractions is
`1/(x^2-4)=1/((x-2)(x+2))=A/(x-2)+B/(x+2)`

`=(A(x+2)+B(x-2))/((x-2)(x+2)`

So equalising LHS and RHS
`1=A(x+2)+B(x-2)`

If `x=2` then `1=4A` `=>` `A=1/4`
and `x=-2` then `1=-4B` `=>` `B=-1/4`

so we have
`1/(x^2-4)=(1/4)/(x-2)+(-1/4)/(x+2)=1/4(1/(x-2)-1/(x+2))`

`int(dx)/(x^2-4)=1/4(int(dx)/(x-2)-int(dx)/(x+2))`

`=1/4(ln(x-2)-ln(x+2))+C`