# How do you integrate int x^3sqrt(1+x^2) using integration by parts?

Nov 8, 2016

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

#### Explanation:

$I = \int {x}^{3} \sqrt{1 + {x}^{2}} \mathrm{dx}$

Before trying any integration by parts, we should attempt more basic substitutions to simplify the integral. Let $u = 1 + {x}^{2}$. This implies that $\mathrm{du} = 2 x \mathrm{dx}$ and ${x}^{2} = u - 1$.

Rearranging the integral:

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

So, we see that integration by parts won't be necessary at all!

$I = \frac{1}{2} \int \left({u}^{\frac{3}{2}} - {u}^{\frac{1}{2}}\right) \mathrm{du} = \frac{1}{2} \left({u}^{\frac{5}{2}} / \left(\frac{5}{2}\right) - {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right)$

$\textcolor{w h i t e}{I} = \frac{1}{2} \left(\frac{2}{5} {u}^{\frac{5}{2}} - \frac{2}{3} {u}^{\frac{3}{2}}\right) = {u}^{\frac{5}{2}} / 5 - {u}^{\frac{3}{2}} / 3 = \frac{3 {u}^{\frac{5}{2}} - 5 {u}^{\frac{3}{2}}}{15}$

$\textcolor{w h i t e}{I} = \frac{{u}^{\frac{3}{2}} \left(3 u - 5\right)}{15} = \frac{{\left(1 + {x}^{2}\right)}^{\frac{3}{2}} \left(3 {x}^{2} - 2\right)}{15} + C$