integral of \sqrt[3]{t}(5t^2-3t+2)

解

\int 3 t (5 t^{2} - 3 t + 2) d t

\frac{3}{2} t^{\frac{10}{3}} - \frac{9}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}} + C

解答ステップ

\int 3 t (5 t^{2} - 3 t + 2) d t

拡張

3 t (5 t^{2} - 3 t + 2) : 5 t^{\frac{7}{3}} - 3 t^{\frac{4}{3}} + 2 3 t

= \int 5 t^{\frac{7}{3}} - 3 t^{\frac{4}{3}} + 2 3 t d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 5 t^{\frac{7}{3}} d t - \int 3 t^{\frac{4}{3}} d t + \int 2 3 t d t

\int 5 t^{\frac{7}{3}} d t = \frac{3}{2} t^{\frac{10}{3}}

\int 3 t^{\frac{4}{3}} d t = \frac{9}{7} t^{\frac{7}{3}}

\int 2 3 t d t = \frac{3}{2} t^{\frac{4}{3}}

= \frac{3}{2} t^{\frac{10}{3}} - \frac{9}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}}

定数を解答に追加する

= \frac{3}{2} t^{\frac{10}{3}} - \frac{9}{7} t^{\frac{7}{3}} + \frac{3}{2} t^{\frac{4}{3}} + C