How do you integrate $\int \frac{x}{{(x^{2} + 1)}^{2}}$ $int x/((x^2+1)^2)$ 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

$\int \frac{x}{{(x^{2} + 1)}^{2}} d x = - \frac{1}{2 (x^{2} + 1)} + C$ $intx/(x^2+1)^2dx=-1/(2(x^2+1))+C$

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

Let $u = x^{2} + 1$ $u = x^2 + 1$ . Then $d u = 2 x d x$ $du = 2xdx$ and so we have

$\int \frac{x}{{(x^{2} + 1)}^{2}} d x = \frac{1}{2} \int \frac{1}{{(x^{2} + 1)}^{2}} \cdot 2 x d x$ $intx/(x^2+1)^2dx = 1/2int1/(x^2+1)^2*2xdx$

$= \frac{1}{2} \int \frac{1}{u^{2}} d u$ $=1/2int1/u^2du$

$= \frac{1}{2} (- \frac{1}{u} + C)$ $=1/2(-1/u + C)$

$= - \frac{1}{2 u} + C$ $=-1/(2u)+C$

$= - \frac{1}{2 (x^{2} + 1)} + C$ $=-1/(2(x^2+1))+C$

How do you integrate int x/((x^2+1)^2)∫x(x2+1)2int x/((x^2+1)^2) using partial fractions?