# What is the antiderivative of  (x^2-1) / (x^2+1)?

##### 1 Answer
Dec 20, 2016

$x - 2 {\tan}^{-} 1 x + C$

#### Explanation:

$\int \frac{{x}^{2} - 1}{{x}^{2} + 1} \mathrm{dx}$
$= \int \frac{{x}^{2} + 1 - 2}{{x}^{2} + 1} \mathrm{dx}$
$= \int 1 - \frac{2}{1 + {x}^{2}} \mathrm{dx}$
$= x - 2 {\tan}^{-} 1 x + C$.

(If the quoting standard integral $\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = {\tan}^{-} 1 x$ is not acceptable, substitute $x = \tan u$, $\mathrm{dx} = {\sec}^{2} u \mathrm{du}$ and use ${\sec}^{2} u = 1 + {\tan}^{2} u$ to reduce the integral to $\int 1 \mathrm{du}$.)