# Evaluate the following integral  x^3sqrt(1+x^2)?

Feb 26, 2017

To integrate by $u$-substitution see below.

#### Explanation:

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

Let $u = 1 + {x}^{2}$.

This make $\mathrm{du} = 2 x \mathrm{dx}$ and ${x}^{2} = u - 1$

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

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

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

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

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

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

Feb 26, 2017

To integrate by parts, see below.

#### Explanation:

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

We can integrate $t = {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$ by substitution if we think of one of the $x$'s as attached to $\mathrm{dx}$ to be $\mathrm{dt} = 2 x \mathrm{dx}$.

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

Let $u = {x}^{2}$ and $\mathrm{dv} = {\left(1 + {x}^{2}\right)}^{\frac{1}{2}} x \mathrm{dx}$,

so $\mathrm{du} = 2 x \mathrm{dx}$ and $v = \frac{1}{3} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}$

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

We can evaluate the integral by substitution, to get

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

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

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

Feb 26, 2017

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

#### Explanation:

Let me try another approach:

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

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} {x}^{n} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}} = n {x}^{n - 1} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}} + 3 {x}^{n + 1} {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} {x}^{n} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}} = \left(n {x}^{n - 1} \left(1 + {x}^{2}\right) + 3 {x}^{n + 1}\right) {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} {x}^{n} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}} = \left(\left(n + 3\right) {x}^{n + 1} + n {x}^{n - 1}\right) {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

In particular, putting $n = 2$ then $n = 0$ we find:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{5} {x}^{2} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}\right) = \left({x}^{3} + \frac{2}{5} x\right) {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

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

Hence:

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

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