# How do you integrate int (2x^3 -x^2)/((x^2 +1)^2) using partial fractions?

Sep 13, 2017

$\int \frac{\left(2 {x}^{3} - {x}^{2}\right) \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$

=$\int \frac{2 {x}^{2} - x}{4} \cdot \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$

=$\frac{2 {x}^{2} - x}{4} \cdot - \frac{1}{{x}^{2} + 1}$-$\int \left[\frac{4 x - 1}{4} \cdot \mathrm{dx}\right] \cdot \frac{- 1}{{x}^{2} + 1}$

=$- \frac{2 {x}^{2} - x}{4 {x}^{2} + 4}$+$\frac{1}{4}$*$\int \frac{\left(4 x - 1\right) \cdot \mathrm{dx}}{{x}^{2} + 1}$

=$- \frac{2 {x}^{2} - x}{4 {x}^{2} + 4}$+$\frac{1}{2}$$\int \frac{2 x \cdot \mathrm{dx}}{{x}^{2} + 1}$-$\frac{1}{4}$$\int \frac{\mathrm{dx}}{{x}^{2} + 1}$

=$- \frac{2 {x}^{2} - x}{4 {x}^{2} + 4} + \frac{1}{2} \ln \left({x}^{2} + 1\right) - \frac{1}{4} \cdot \arctan x + C$

#### Explanation:

I used integration by parts with $u = \frac{2 {x}^{2} - x}{4}$ and $\mathrm{dv} = \frac{4 x \cdot \mathrm{dx}}{{x}^{2} + 1} ^ 2$