다음 적분을 구하라.
$$\int_2^3\frac{t^3}{t+1}dt+\int_2^3\frac{1}{u+1}du$$
풀이
\(\displaystyle\int_2^3\frac{t^3}{t+1}dt+\int_2^3\frac{1}{u+1}du\)
\(\displaystyle=\int_2^3\frac{(t+1)(t^2-t+1)-1}{t+1}dt+\int_2^3\frac{1}{u+1}du\)
\(\displaystyle=\int_2^3\frac{(t+1)(t^2-t+1)}{t+1}dt-\int_2^3\frac{1}{t+1}dt+\int_2^3\frac{1}{u+1}du\)
\(\displaystyle=\int_2^3(t^2-t+1)dt=\left[\frac{t^3}{3}-\frac{t^2}{2}+t\right]_2^3\)
\(\displaystyle=\frac{3^3}{3}-\frac{3^2}{2}+3-\left(\frac{2^3}{3}-\frac{2^2}{2}+2\right)\)
\(\displaystyle=9-\frac{9}{2}+3-\left(\frac{8}{3}-2+2\right)\)
\(\displaystyle=\frac{15}{2}-\frac{8}{3}=\frac{45-16}{6}=\frac{29}{6}\)