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

解

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

\frac{6}{11} x^{\frac{11}{6}} - \frac{2}{3} x^{\frac{3}{2}} + \frac{3}{4} x^{\frac{4}{3}} - x + C

解答ステップ

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

拡張

(x + 1) (3 x - 1) : x^{\frac{5}{6}} - x + 3 x - 1

= \int x^{\frac{5}{6}} - x + 3 x - 1 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^{\frac{5}{6}} d x - \int x d x + \int 3 x d x - \int 1 d x

\int x^{\frac{5}{6}} d x = \frac{6}{11} x^{\frac{11}{6}}

\int x d x = \frac{2}{3} x^{\frac{3}{2}}

\int 3 x d x = \frac{3}{4} x^{\frac{4}{3}}

\int 1 d x = x

= \frac{6}{11} x^{\frac{11}{6}} - \frac{2}{3} x^{\frac{3}{2}} + \frac{3}{4} x^{\frac{4}{3}} - x

定数を解答に追加する

= \frac{6}{11} x^{\frac{11}{6}} - \frac{2}{3} x^{\frac{3}{2}} + \frac{3}{4} x^{\frac{4}{3}} - x + C