∫ ( ( 1 − t ) ( 2 + t 2 ) ) d t = ∫ ( 2 + t 2 − 2 t − t 3 ) d t = 2 t + t 3 3 − 2 t 2 2 − t 4 4 + C = 2 t − t 2 + t 3 3 − t 4 4 + C {\displaystyle {\begin{aligned}\int \left((1-t)(2+t^{2})\right)dt=\int (2+t^{2}-2t-t^{3})dt=2t+{\frac {t^{3}}{3}}-{\frac {2t^{2}}{2}}-{\frac {t^{4}}{4}}+C=2t-t^{2}+{\frac {t^{3}}{3}}-{\frac {t^{4}}{4}}+C\end{aligned}}}
