integral of (5x+2)/(\sqrt[3]{6x+3)}

解

\int \frac{5 x + 2}{3 6 x + 3} d x

\frac{3 ^{\frac{2}{3}}}{8} (4 x - 3) (2 x + 1)^{\frac{2}{3}} + \frac{1}{2} (6 x + 3)^{\frac{2}{3}} + C

解答ステップ

\int \frac{5 x + 2}{3 6 x + 3} d x

拡張

\frac{5 x + 2}{3 6 x + 3} : \frac{5 x}{3 6 x + 3} + \frac{2}{3 6 x + 3}

= \int \frac{5 x}{3 6 x + 3} + \frac{2}{3 6 x + 3} 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{5 x}{3 6 x + 3} d x + \int \frac{2}{3 6 x + 3} d x

\int \frac{5 x}{3 6 x + 3} d x = \frac{3 ^{\frac{2}{3}}}{8} (4 x - 3) (2 x + 1)^{\frac{2}{3}}

\int \frac{2}{3 6 x + 3} d x = \frac{1}{2} (6 x + 3)^{\frac{2}{3}}

= \frac{3 ^{\frac{2}{3}}}{8} (4 x - 3) (2 x + 1)^{\frac{2}{3}} + \frac{1}{2} (6 x + 3)^{\frac{2}{3}}

定数を解答に追加する

= \frac{3 ^{\frac{2}{3}}}{8} (4 x - 3) (2 x + 1)^{\frac{2}{3}} + \frac{1}{2} (6 x + 3)^{\frac{2}{3}} + C