# How do you integrate int -x/((x+1)-sqrt(x+1))dx?

Feb 20, 2017

$- x - 2 \sqrt{x + 1} + C$

#### Explanation:

In order to integrate this, we need to use a method called u-substitution, but a good idea is to remove all constants out of the integral first.

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

Let $u = \sqrt{x + 1}$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x + 1}}$

$\mathrm{dx} = 2 \sqrt{x + 1} \times \mathrm{du} = 2 u$ $\mathrm{du}$

$x = {u}^{2} - 1$

The integral can now be rewritten in terms of $u$

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

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

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

Since $1$ is a constant, we can add it to $C$, leaving the answer as:

$- x - 2 \sqrt{x + 1} + C$