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

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Answer 1 · 2016-02-21T00:23:22

Rather than using partial fractions, we can make a simple substitution to find that

#intx/(x^2+1)^2dx=-1/(2(x^2+1))+C#

Rather than partial fractions, this is easiest to solve through integration by substitution.

Let #u = x^2 + 1#. Then #du = 2xdx# and so we have

#intx/(x^2+1)^2dx = 1/2int1/(x^2+1)^2*2xdx#

#=1/2int1/u^2du#

#=1/2(-1/u + C)#

#=-1/(2u)+C#

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