integral of (x+1)\sqrt[3]{4-x}

解

\int (x + 1) 3 4 - x d x

\frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}} + C

解答ステップ

\int (x + 1) 3 4 - x d x

拡張

(x + 1) 3 4 - x : x 3 4 - x + 3 4 - x

= \int x 3 4 - x + 3 4 - x d x

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 x 3 4 - x d x + \int 3 4 - x d x

\int x 3 4 - x d x = \frac{3}{7} (4 - x)^{\frac{7}{3}} - 3 (4 - x)^{\frac{4}{3}}

\int 3 4 - x d x = - \frac{3}{4} (4 - x)^{\frac{4}{3}}

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - 3 (4 - x)^{\frac{4}{3}} - \frac{3}{4} (4 - x)^{\frac{4}{3}}

類似した元を足す:

- 3 (- x + 4)^{\frac{4}{3}} - \frac{3}{4} (- x + 4)^{\frac{4}{3}} = - \frac{15}{4} (- x + 4)^{\frac{4}{3}}

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}}

定数を解答に追加する

= \frac{3}{7} (4 - x)^{\frac{7}{3}} - \frac{15}{4} (4 - x)^{\frac{4}{3}} + C