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

1 Answer
Jan 24, 2017

# 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)) #