# How do you integrate int xsqrt(3x+ 1) dx?

##### 1 Answer
Apr 15, 2017

$\int x \sqrt{3 x + 1} = \frac{2}{9} x {\left(3 x + 1\right)}^{\frac{3}{2}} - \frac{4}{135} {\left(3 x + 1\right)}^{\frac{5}{2}} + C$

#### Explanation:

You will need substitution in addition to integration by parts. Let $u = x$ and $\mathrm{dv} = \sqrt{3 x + 1}$. By the power rule, $\mathrm{du} = \mathrm{dx}$. But we will need to make a substitution to find $v$. Let $t = 3 x + 1$. Then $\mathrm{dt} = 3 \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{dt}}{3}$.

$\int \sqrt{3 x + 1} = \sqrt{t} \cdot \frac{\mathrm{dt}}{3} = \frac{1}{3} \sqrt{t} \mathrm{dt} = \frac{2}{9} {t}^{\frac{3}{2}} = \frac{2}{9} {\left(3 x + 1\right)}^{\frac{3}{2}}$

Apply integration by parts now.

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

$\int x \sqrt{3 x + 1} = \frac{2}{9} x {\left(3 x + 1\right)}^{\frac{3}{2}} - \int \frac{2}{9} {\left(3 x + 1\right)}^{\frac{3}{2}} \mathrm{dx}$

$\int x \sqrt{3 x + 1} = \frac{2}{9} x {\left(3 x + 1\right)}^{\frac{3}{2}} - \frac{2}{9} \int {\left(3 x + 1\right)}^{\frac{3}{2}} \mathrm{dx}$

Now let $n = 3 x + 1$. Then $\mathrm{dn} = 3 \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{dn}}{3}$.

$\int {\left(3 x + 1\right)}^{\frac{3}{2}} = \frac{1}{3} \int {n}^{\frac{3}{2}} \mathrm{dn} = \frac{2}{15} {n}^{\frac{5}{2}} = \frac{2}{15} {\left(3 x + 1\right)}^{\frac{5}{2}}$

$\int x \sqrt{3 x + 1} = \frac{2}{9} x {\left(3 x + 1\right)}^{\frac{3}{2}} - \frac{2}{9} \left(\frac{2}{15}\right) {\left(3 x + 1\right)}^{\frac{5}{2}} + C$

$\int x \sqrt{3 x + 1} = \frac{2}{9} x {\left(3 x + 1\right)}^{\frac{3}{2}} - \frac{4}{135} {\left(3 x + 1\right)}^{\frac{5}{2}} + C$

Hopefully this helps!