How do you integrate #int (x+4) / [(x-1)(x^2+4)]# using partial fractions?
1 Answer
Nov 23, 2016
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
Put
Compare coefficients of
Hence:
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 #