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

Question

1664 views around the world

You can reuse this answer
Creative Commons License

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

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

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

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

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

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

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