# How do you integrate int dx/sqrt(x^2+2x) using trig substitutions?

Mar 29, 2018

The answer is $= \ln \left(| \sqrt{{x}^{2} + 2 x} + x + 1 |\right) + C$

#### Explanation:

Complete the square in the denominator

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

Let $x + 1 = \sec u$, $\implies$, $\mathrm{dx} = \sec u \tan u \mathrm{du}$

Therefore, the integral is

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

$= \int \frac{\sec u \tan u \mathrm{du}}{\sqrt{{\tan}^{2} u}}$

$= \int \left(\sec u \mathrm{du}\right)$

$= \int \frac{\sec u \left(\tan u + \sec u\right) \mathrm{du}}{\tan u + \sec u}$

Let $v = \tan u + \sec u$, $\implies$, $\mathrm{dv} = \left(\sec u \tan u + {\sec}^{2} u\right) \mathrm{du}$

Therefore,

$I = \int \frac{\mathrm{dv}}{v}$

$= \ln v$

$= \ln \left(\tan u + \sec u\right)$

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

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