integral of (x-5)/(\sqrt[3]{2x+1)}

解

\int \frac{x - 5}{3 2 x + 1} d x

\frac{1}{4} (\frac{3}{5} (2 x + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x + 1)^{\frac{2}{3}}) - \frac{15}{4} (2 x + 1)^{\frac{2}{3}} + C

解答ステップ

\int \frac{x - 5}{3 2 x + 1} d x

拡張

\frac{x - 5}{3 2 x + 1} : \frac{x}{3 2 x + 1} - \frac{5}{3 2 x + 1}

= \int \frac{x}{3 2 x + 1} - \frac{5}{3 2 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 \frac{x}{3 2 x + 1} d x - \int \frac{5}{3 2 x + 1} d x

\int \frac{x}{3 2 x + 1} d x = \frac{1}{4} (\frac{3}{5} (2 x + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x + 1)^{\frac{2}{3}})

\int \frac{5}{3 2 x + 1} d x = \frac{15}{4} (2 x + 1)^{\frac{2}{3}}

= \frac{1}{4} (\frac{3}{5} (2 x + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x + 1)^{\frac{2}{3}}) - \frac{15}{4} (2 x + 1)^{\frac{2}{3}}

定数を解答に追加する

= \frac{1}{4} (\frac{3}{5} (2 x + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x + 1)^{\frac{2}{3}}) - \frac{15}{4} (2 x + 1)^{\frac{2}{3}} + C