# How do you integrate x^2/(x^2+1)?

Jun 28, 2016

$x - \arctan x + C$

#### Explanation:

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

$\int \setminus 1 - \frac{1}{{x}^{2} + 1} \setminus \mathrm{dx}$

$= x - \textcolor{red}{\int \setminus \frac{1}{{x}^{2} + 1} \setminus \mathrm{dx}}$

in terms of the red bit, use sub $x = \tan t , \mathrm{dx} = {\sec}^{2} t \setminus \mathrm{dt}$

this makes it

$\setminus \int \setminus \frac{1}{{\tan}^{2} t + 1} \setminus {\sec}^{2} t \setminus \mathrm{dt}$

$= \setminus \int \setminus \frac{1}{{\sec}^{2} t} \setminus {\sec}^{2} t \setminus \mathrm{dt}$

$= \setminus \int \setminus \mathrm{dt}$

$= \arctan x - C$

So the full integral is

$x - \arctan x + C$