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

Feb 7, 2016

$2 {\tan}^{-} 1 \left(x\right) + C$

#### Explanation:

Normally this may be quoted as a standard integral:

$\int \frac{2}{{x}^{2} + 1} \mathrm{dx} = 2 \int \frac{1}{{x}^{2} + 1} \mathrm{dx} = 2 {\tan}^{-} 1 \left(x\right) + C$

However to convince yourself this is indeed the case:

First consider the trig - identity:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

Divide this through by ${\cos}^{2} \left(x\right)$ to obtain the identity:

${\sin}^{2} \frac{x}{\cos} ^ 2 \left(x\right) + {\cos}^{2} \frac{x}{\cos} ^ 2 \left(x\right) = \frac{1}{\cos} ^ 2 \left(x\right) \to {\tan}^{2} \left(x\right) + 1 = {\sec}^{2} \left(x\right)$

Now going back to the integral, use the substitution:

$\tan \left(u\right) = x$

This will also mean:

${\sec}^{2} \left(u\right) \mathrm{du} = \mathrm{dx}$ Now put this substitution into the integral:

$\int \frac{2}{{x}^{2} + 1} \mathrm{dx} = 2 \int {\sec}^{2} \frac{u}{{\tan}^{2} \left(u\right) + 1} \mathrm{du}$

Now using the trig-identity we just saw we can replace the denominator giving us:

$2 \int {\sec}^{2} \frac{u}{\sec} ^ 2 \left(u\right) \mathrm{du} = 2 \int 1 \mathrm{du} = u + C$

Now reverse the substitution and we get:

$= 2 {\tan}^{-} 1 \left(x\right) + C$