# How do you integrate x^5 * sqrt(x^3+1) dx?

May 2, 2017

The integral is $\frac{2}{15} {\left({x}^{3} + 1\right)}^{\frac{5}{2}} - \frac{2}{9} {\left({x}^{3} + 1\right)}^{\frac{3}{2}} + C$

#### Explanation:

Let $u = {x}^{3} + 1$. Then $\mathrm{du} = 3 {x}^{2} \mathrm{dx} \to \mathrm{dx} = \frac{\mathrm{du}}{3 {x}^{2}}$. Let $I$ be the integral.

$I = \int {x}^{5} \sqrt{u} \cdot \frac{\mathrm{du}}{3 {x}^{2}}$

$I = \frac{1}{3} \int {x}^{3} \sqrt{u} \mathrm{du}$

$I = \frac{1}{3} \int \left(u - 1\right) \sqrt{u} \mathrm{du}$

$I = \frac{1}{3} \int \left(u - 1\right) {u}^{\frac{1}{2}} \mathrm{du}$

$I = \frac{1}{3} \int {u}^{\frac{3}{2}} - {u}^{\frac{1}{2}} \mathrm{du}$

$I = \frac{1}{3} \int {u}^{\frac{3}{2}} \mathrm{du} - \frac{1}{3} \int {u}^{\frac{1}{2}} \mathrm{du}$

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

$I = \frac{2}{15} {u}^{\frac{5}{2}} - \frac{2}{9} {u}^{\frac{3}{2}} + C$

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

Hopefully this helps!