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

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Answer 1 · 2018-05-25T15:58:51

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

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#