Question

How do you use partial fractions to find the integral `int 1/(x^2-4)dx`?

Answer 1

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

Explanation:

We can factorise the denominator of the integrand as follows:

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

And we can decompose this into partial fractions of the form;

`1/((x+2)(x-2)) -= A/(x+2)+B/(x-2)`
`" " = (A(x-2)+B(x+2)) / ((x+2)(x-2))`
`:. " " 1 = A(x-2)+B(x+2)`

Put `x=-2 => 1 =-4A => A=-1/4`
Put `x=2 \ \ \ \ \ => 1 =4B \ \ \ \ => B=1/4`

Hence the partial fraction decomposition is:

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

And so the integral can be written;

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

Which we can now easily integrate to get;

`int \ 1/(x^2-4) \ dx = 1/4 ln abs(x-2) - 1/4ln abs(x+2) + C`
`" " = 1/4 ln abs(x-2) - 1/4ln abs(x+2) + 1/4lnA`
`" " = 1/4 ln Aabs((x-2)/(x+2))`