Compute this integral: $$\int _1^8\left(\sqrt[3]{r}+\frac{1}{\sqrt[3]{r}}\right)\:dr$$
My attempt:
$$\int \sqrt[3]{r}+\frac{1}{\sqrt[3]{r}}dr$$
$$\int \sqrt[3]{r}dr=\frac{3}{4}r^{\frac{4}{3}}$$
$$=\frac{3}{4}r^{\frac{4}{3}}+\frac{3}{2}r^{\frac{2}{3}}+C$$
$$\quad \int _1^8\sqrt[3]{r}+\frac{1}{\sqrt[3]{r}}dr=18-\frac{9}{4}$$
$$\lim _{r\to \:1+}\left(\frac{3}{4}r^{\frac{4}{3}}+\frac{3}{2}r^{\frac{2}{3}}\right)=\frac{9}{4}$$
$$\lim _{r\to \:8-}\left(\frac{3}{4}r^{\frac{4}{3}}+\frac{3}{2}r^{\frac{2}{3}}\right)=18$$
$$=18-\frac{9}{4}$$
$$=\frac{63}{4}$$
Is this correct?
I believe it is more efficient to get rid of the cube root by enforcing the substitution $r=x^3$ in the very first stage: $$ \int_{1}^{8}\left(\frac{1}{\sqrt[3]{r}}+\sqrt[3]{r}\right)\,dr = \int_{1}^{2}3x^2\left(\frac{1}{x}+x\right)\,dx = 3\left[\frac{x^2}{2}+\frac{x^4}{4}\right]_{1}^{2}=\frac{63}{4}.\quad\color{green}{\checkmark} $$