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

解答

\frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}} + C

求解步骤

使用换元积分法

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

乘开

使用积分加法定则:

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

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

= \frac{1}{12} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

u = 2 x^{3} + 1

代回

= \frac{1}{12} (\frac{3}{5} (2 x^{3} + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x^{3} + 1)^{\frac{2}{3}})

化简

\frac{1}{12} (\frac{3}{5} (2 x^{3} + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x^{3} + 1)^{\frac{2}{3}}) : \frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}}

= \frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}}

解答补常数

= \frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}} + C