∫ − 2 2 ( 3 u + 1 ) 2 d u ∫ 3 u 2 + 6 u + 1 d u {\displaystyle {\begin{aligned}&\int _{-2}^{2}({3u+1})^{2}du&\int _{3u^{2}+6u+1}du\end{aligned}}}
