# How do you find the antiderivative of int 1/sqrt(1+x^2) dx?

Jun 23, 2017

$\ln \left\mid x + \sqrt{1 + {x}^{2}} \right\mid + C$

#### Explanation:

$I = \int \frac{1}{\sqrt{1 + {x}^{2}}} \mathrm{dx}$

Let $x = \tan \theta$. This implies that $\mathrm{dx} = {\sec}^{2} \theta d \theta$.

$I = \int \frac{1}{\sqrt{1 + {\tan}^{2} \theta}} {\sec}^{2} \theta d \theta$

Since $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$:

$I = \int \sec \theta d \theta = \ln \left\mid \sec \theta + \tan \theta \right\mid$

Note that $\tan \theta = x$ and $\sec \theta = \sqrt{1 + {\tan}^{2} \theta} = \sqrt{1 + {x}^{2}}$:

$I = \ln \left\mid x + \sqrt{1 + {x}^{2}} \right\mid + C$