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

Question

3499 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-24T02:19:24

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

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;

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

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