Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 30

Evaluate $\displaystyle \int^1_0 \frac{r^3}{\sqrt{4 + r^2}} dr$ by using Integration by parts.
If we let $u = r^2$ and $\displaystyle dv = \frac{r dr}{\sqrt{4+r^2}}$, then
$du - 2r d$ and $\displaystyle v = \int \frac{rdr}{\sqrt{4+r^2}}$

To evaluate $\displaystyle \int \frac{rdr}{\sqrt{4+r^2}}$, we let $u_1 = 4 + r^2$, then $du_1 = 2rdr$

So,

$\begin{equation} \begin{aligned} \int \frac{rdr}{\sqrt{4+r^2}} = \int \frac{du_1}{2} \left( \frac{1}{\sqrt{u_1}} \right) &= \frac{1}{2} \int \left( u_1^{-\frac{1}{2}} \right) du_1\\ \\ &= \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}}\\ \\ &= \sqrt{4+r^2} \end{aligned} \end{equation}$


Hence, from integration by parts

$\begin{equation} \begin{aligned} \int^1_0 \frac{r^3}{\sqrt{4+r^2}} dr = uv - \int v du &= r^2 \sqrt{4+r^2} - \int \left( \sqrt{4+r^2}\right)(2r dr)\\ \\ &= r^2 \sqrt{4+r^2} -2 \int r \sqrt{4+r^2} dr \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{if we let } u_2 &= 4 + r^2 \text{ , then}\\ \\ du_2 &= 2r dr \end{aligned} \end{equation}$


So,

$\begin{equation} \begin{aligned} \int r \sqrt{4 + r^2} dr = \int \sqrt{u_2} \left( \frac{du_2}{2} \right) &= \frac{1}{2} \int (u_2)^{\frac{1}{2}} du_2\\ \\ &= \frac{1}{2} \frac{(u)^{\frac{3}{2}}}{\frac{3}{2}}\\ \\ &= \frac{(4+r^2)^{\frac{3}{2}}}{3} \end{aligned} \end{equation}$


Therefore,
$\displaystyle \int^1_0 \frac{r^3}{\sqrt{4+r^2}} dr = r^2 \sqrt{4 + r^2} - 2 \left[ \frac{(4+r^2)^{\frac{3}{2}}}{3} \right]$
Evaluating from $x = 0$ to $x = 1$,
$\displaystyle \int^1_0 \frac{r^3}{\sqrt{4 + r^2}} = \frac{16-7\sqrt{5}}{3}$