# How do you evaluate the integral int 1/(xsqrt(2+x^2))?

Jan 15, 2017

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

#### Explanation:

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

Use the substitution $x = \sqrt{2} \tan \theta$. This implies that ${x}^{2} = 2 {\tan}^{2} \theta$ and $\mathrm{dx} = \sqrt{2} {\sec}^{2} \theta d \theta$.

Plugging this in shows that:

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

Note that $\sqrt{2 + 2 {\tan}^{2} \theta} = \sqrt{2} \sqrt{1 + {\tan}^{2} \theta} = \sqrt{2} \sec \theta$. This comes from the identity $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$.

$I = \int \frac{\sqrt{2} {\sec}^{2} \theta d \theta}{\sqrt{2} \tan \theta \left(\sqrt{2} \sec \theta\right)} = \frac{1}{\sqrt{2}} \int \sec \frac{\theta}{\tan} \theta d \theta$

Rewriting with $\sin \theta$ and $\cos \theta$:

$I = \frac{1}{\sqrt{2}} \int \frac{1}{\cos} \theta \cos \frac{\theta}{\sin} \theta d \theta = \frac{1}{\sqrt{2}} \int \csc \theta d \theta = - \frac{1}{\sqrt{2}} \ln \left\mid \csc \theta + \cot \theta \right\mid$

From the original substitution we have $\tan \theta = \frac{x}{\sqrt{2}}$. This correlates to a right triangle with the side opposite $\theta$ being $x$ and the side adjacent to $\theta$ being $\sqrt{2}$. The Pythagorean theorem gives the hypotenuse as $\sqrt{{x}^{2} + 2}$.

Thus, $\csc \theta$ is the hypotenuse over the opposite side, or $\csc \theta = \frac{\sqrt{{x}^{2} + 2}}{x}$ and $\cot \theta = \frac{1}{\tan} \theta = \frac{\sqrt{2}}{x}$.

So:

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

We can rewrite this using $B \ln \left(A\right) = \ln \left({A}^{B}\right)$. In this case, we will use $- \ln \left(A\right) = \ln \left({A}^{-} 1\right) = \ln \left(\frac{1}{A}\right)$.

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