# Question #6f533

Dec 15, 2016

$\int x \sqrt{2 x + 1} \mathrm{dx} = {\left(2 x + 1\right)}^{\frac{3}{2}} \left(\frac{1}{5} x - \frac{1}{15}\right) + C$

#### Explanation:

Using the integration by parts formula

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let
$u = x \implies \mathrm{du} = \mathrm{dx}$
$\mathrm{dv} = {\left(2 x + 1\right)}^{\frac{1}{2}} \mathrm{dx} \implies v = \frac{1}{3} {\left(2 x + 1\right)}^{\frac{3}{2}}$

$\int x \sqrt{2 x + 1} \mathrm{dx} = \int u \mathrm{dv}$

$= u v - \int v \mathrm{du}$

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

$= \frac{1}{3} x {\left(2 x + 1\right)}^{\frac{3}{2}} - \frac{1}{3} \left[\frac{1}{5} {\left(2 x + 1\right)}^{\frac{5}{2}}\right] + C$

$= \frac{1}{3} x {\left(2 x + 1\right)}^{\frac{3}{2}} - \frac{1}{15} {\left(2 x + 1\right)}^{\frac{5}{2}} + C$

$= \frac{1}{3} {\left(2 x + 1\right)}^{\frac{3}{2}} \left[x - \frac{1}{5} \left(2 x + 1\right)\right] + C$

$= {\left(2 x + 1\right)}^{\frac{3}{2}} \left(\frac{1}{5} x - \frac{1}{15}\right) + C$