다음 적분을 구하라.
$$\int_1^3\frac{u^4-8}{u^2}du$$
풀이
\(\displaystyle\int_1^3\frac{u^4-8}{u^2}du=\int_1^3\left(u^2-8u^{-2}\right)du\)
\(\displaystyle=\left[\frac{1}{3}u^3-8(-1)u^{-1}\right]_1^3=\left[\frac{u^3}{3}+\frac{8}{u}\right]_1^3\)
\(\displaystyle=\frac{3^3}{3}+\frac{8}{3}-\left(\frac{1^3}{3}+\frac{8}{1}\right)=\frac{35}{3}-\frac{25}{3}=\frac{10}{3}\)