# How do you integrate int x/sqrt(x+1) using substitution?

Oct 24, 2016

$y = \frac{2 {\left(x + 1\right)}^{\frac{3}{2}}}{3} - 2 \sqrt{x + 1} + c$

#### Explanation:

so let us take,

$u = x + 1$

so we differentiate this to find that

$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

and $x = u - 1$

so let's rewrite our original integral

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

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

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

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

$= \frac{2 {u}^{\frac{3}{2}}}{3} - 2 \sqrt{u} + c$

we can now sub in the $x + 1 \text{ for the u's}$ if we need,

$= \frac{2 {\left(x + 1\right)}^{\frac{3}{2}}}{3} - 2 \sqrt{x + 1} + c$

Oct 24, 2016

The integral is $= \frac{2}{3} \sqrt{x + 1} \left(x - 2\right) + C$

#### Explanation:

Let $u = x + 1$ and $x = u - 1$

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

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

$= {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)$

$= \frac{2}{3} u \cdot \sqrt{u} - 2 \sqrt{u}$
$= 2 \cdot \sqrt{u} \left(\frac{1}{3} u - 1\right)$
$= \frac{2}{3} \sqrt{u} \left(u - 3\right)$

Replacing u by (x+1)

$\int \frac{x \mathrm{dx}}{\sqrt{x + 1}} = \frac{2}{3} \sqrt{x + 1} \left(x + 1 - 3\right) + C$

$= \frac{2}{3} \sqrt{x + 1} \left(x - 2\right) + C$