How do you integrate #int (x+4) / [(x-1)(x^2+4)]# using partial fractions?

1 Answer
Nov 23, 2016

# int (x+4)/((x-1)(x^2+4)) = ln|x-1| - 1/2ln(x^2+4) +c #

Explanation:

The Partial fraction decomposition of the integrand will be of the form;

# (x+4)/((x-1)(x^2+4)) -= A/(x-1) + (Bx+C)/(x^2+4) #
# :. (x+4)/((x-1)(x^2+4)) = (A(x^2+4) + (Bx+C)(x-1))/((x-1)(x^2+4)) #
# :. (x+4) -= A(x^2+4) + (Bx+C)(x-1) #

Put #x=1 => 5 = A(1+4) +0 #
# :. 5A=5 => A=1 #

Put #x=0 => 4=4A+(C)(-1) #
# :. C=4-4= 0#

Compare coefficients of #x^2 => 0 = A+B#
# :. B=-1 #

Hence:
# (x+4)/((x-1)(x^2+4)) = 1/(x-1) -x/(x^2+4) #

And so,

# int (x+4)/((x-1)(x^2+4)) = int 1/(x-1) - int x/(x^2+4) #
# :. int (x+4)/((x-1)(x^2+4)) = int 1/(x-1) - 1/2int (2x)/(x^2+4) #
# :. int (x+4)/((x-1)(x^2+4)) = ln|x-1| - 1/2ln(x^2+4) +c #