# How do you integrate? int(2x - 2)/(sqrtx +1) dx

## $\int \frac{2 x - 2}{\sqrt{x} + 1} \mathrm{dx}$

Apr 7, 2018

$I = \frac{4}{3} {x}^{\frac{3}{2}} - 2 x + C$

#### Explanation:

We want to integrate

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

Rewrite the integrand

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

$\textcolor{w h i t e}{I} = 2 \int \frac{\left(\sqrt{x} + 1\right) \left(\sqrt{x} - 1\right)}{\sqrt{x} + 1} \mathrm{dx}$

$\textcolor{w h i t e}{I} = 2 \int \sqrt{x} - 1 \mathrm{dx}$

Integrate by the power rule

$I = 2 \left(\frac{2}{3} {x}^{\frac{3}{2}} - x\right) + C$

$\textcolor{w h i t e}{I} = \frac{4}{3} {x}^{\frac{3}{2}} - 2 x + C$

Apr 7, 2018

$I = \frac{4}{3} {x}^{\frac{3}{2}} - 2 x + c$

#### Explanation:

Here,

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

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

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

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

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

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

$= 2 \left[{x}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - x\right] + c$

$= \frac{4}{3} {x}^{\frac{3}{2}} - 2 x + c$