# How do you integrate int 1/sqrt(x^2+2x) by trigonometric substitution?

Feb 20, 2017

$\ln | \sqrt{{\left(x + 1\right)}^{2} - 1} + x + 1 | + C$

#### Explanation:

Complete the square in the denominator (within the sqrt).

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

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

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

Let $u = x + 1$. Then $\mathrm{du} = \mathrm{dx}$.

$\int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

Now use the substitution $u = \sec \theta$. Then $\mathrm{du} = \sec \theta \tan \theta d \theta$.

$\int \frac{1}{\sqrt{{\sec}^{2} \theta - 1}} \sec \theta \tan \theta d \theta$

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

$\int \frac{1}{\tan} \theta \sec \theta \tan \theta d \theta$

$\int \sec \theta d \theta$

This is a known integral that can be derived here

$\ln | \sec \theta + \tan \theta | + C$

Obviously it's not good enough to stay in $\theta$; we have to return to $x$. From our initial substitution, $\frac{u}{1} = \sec \theta$. This means that the side opposite $\theta$ measures $\sqrt{{u}^{2} - 1}$. This also means that $\tan \theta = \frac{\sqrt{{u}^{2} - 1}}{1} = \sqrt{{u}^{2} - 1}$.

$\ln | \sqrt{{u}^{2} - 1} + u | + C$

We have one more substitution to reverse.

$\ln | \sqrt{{\left(x + 1\right)}^{2} - 1} + x + 1 | + C$

Hopefully this helps!