## Calculus: Early Transcendentals 8th Edition

$$\displaystyle\int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\thinspace dr = \frac{16-7\sqrt 5}{3}$$
$\displaystyle\int_{0}^{1}\frac{r^{3}}{\sqrt{4+r^{2}}}\thinspace dr = \displaystyle\int_{0}^{1}\bigg(\frac{r^{2}}{\sqrt{4+r^{2}}}\times r\bigg)\thinspace dr$ $u = 4+r^{2} \rightarrow r^{2} = u-4$ $du = 2rdr \rightarrow \frac{du}{2} = rdr$ Using u substitution gives: $$\displaystyle\int_{0}^{1} \frac{u-4}{\sqrt u}\thinspace \frac{du}{2}$$ $$=\frac{1}{2}\displaystyle\int_{0}^{1} \frac{u}{\sqrt u}-\frac{4}{\sqrt u}\thinspace du$$ $$=\frac{1}{2}\displaystyle\int_{0}^{1} u^{\frac{1}{2}}-4u^{-\frac{1}{2}}\thinspace du$$ Once you find the new variables you can find a new integration interval by this: when $r =1 \rightarrow u = 5$ and when $r = 0 \rightarrow u = 4$; use the $u = 4+r^{2}$ and for every case use the $u$ that fits to your equation. Now the equivalent integral is: $$=\frac{1}{2}\displaystyle\int_{4}^{5} u^{\frac{1}{2}}-4u^{-\frac{1}{2}}\thinspace du$$ $$=\frac{1}{2}\bigg[\frac{2}{3}u^{\frac{3}{2}}-8u^{\frac{1}{2}}\bigg]^{5}_4$$ $$=\frac{1}{2}\bigg(\frac{2}{3} 5^{\frac{3}{2}} - 8\sqrt 5\bigg) - \bigg(\frac{2}{3}\times 8-8\times 2 \bigg)$$ $$=\frac{1}{2} \bigg(\frac{10\sqrt 5}{3} - \frac{24\sqrt 5}{3}-\frac{16}{3}+16\bigg)$$ $$=\frac{5\sqrt 5}{3}-\frac{12\sqrt 5}{3}-\frac{8}{3}+8$$ $$= -\frac{7\sqrt 5}{3} +\frac{16}{3} = \frac{16-7\sqrt 5}{3}$$