20. (int_{0}^{1}\frac{r^{3}}{sqrt{4 + r^{2}}}dr)

Explanation:

Step1: Use substitution

Let $u = r^{2}+4$, then $du = 2r\ dr$. When $r = 0$, $u = 4$; when $r = 1$, $u=5$. And $r^{3}dr=\frac{1}{2}r^{2}\times(2r\ dr)$. Since $r^{2}=u - 4$, the integral $\int_{0}^{1}\frac{r^{3}}{\sqrt{r^{2}+4}}dr=\frac{1}{2}\int_{4}^{5}\frac{u - 4}{\sqrt{u}}du$.

Step2: Split the integral

$\frac{1}{2}\int_{4}^{5}\frac{u - 4}{\sqrt{u}}du=\frac{1}{2}\int_{4}^{5}(\frac{u}{\sqrt{u}}-\frac{4}{\sqrt{u}})du=\frac{1}{2}\int_{4}^{5}(u^{\frac{1}{2}}-4u^{-\frac{1}{2}})du$.

Step3: Integrate term - by - term

$\frac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}-4\times2u^{\frac{1}{2}}\right]{4}^{5}=\frac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}} - 8u^{\frac{1}{2}}\right]{4}^{5}$.

Step4: Evaluate the definite integral

$\frac{1}{2}\left[\left(\frac{2}{3}\times5^{\frac{3}{2}}-8\times5^{\frac{1}{2}}\right)-\left(\frac{2}{3}\times4^{\frac{3}{2}}-8\times4^{\frac{1}{2}}\right)\right]=\frac{1}{2}\left[\frac{2}{3}\times5\sqrt{5}-8\sqrt{5}-\frac{2}{3}\times8 + 16\right]=\frac{1}{2}\left[\left(\frac{10}{3}-8\right)\sqrt{5}-\frac{16}{3}+16\right]=\frac{1}{2}\left[\left(\frac{10 - 24}{3}\right)\sqrt{5}+\frac{- 16 + 48}{3}\right]=\frac{1}{2}\left[-\frac{14}{3}\sqrt{5}+\frac{32}{3}\right]=\frac{16}{3}-\frac{7}{3}\sqrt{5}$.

Answer:

$\frac{16}{3}-\frac{7}{3}\sqrt{5}$