How do you integrate #int-x/(1-x^2)^4dx# using integration of rational functions by partial fractions?

1 Answer
May 25, 2018

The answer is #=-1/(6(1-x^2)^3)+C#

Explanation:

There is no need to perform the decomposition into partial fractions.

Perform this integral by substitution

Let #u=1-x^2#, #=>#, #du=-2xdx#

#=>#, #-xdx=1/2du#

Therefore, the integral is

#int(-xdx)/(1-x^2)^4=1/2int(du)/u^4#

#=1/2(-1/(3u^3))#

#=-1/(6(1-x^2)^3)+C#