integral of x^3\sqrt[3]{8-3x^4}

解答

\int x^{3} 3 8 - 3 x^{4} d x

- \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}} + C

求解步骤

\int x^{3} 3 8 - 3 x^{4} d x

使用换元积分法

= \int - \frac{3 u}{12} d u

提出常数:

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

= - \frac{1}{12} \cdot \int 3 u d u

使用幂法则

= - \frac{1}{12} \cdot \frac{3}{4} u^{\frac{4}{3}}

u = 8 - 3 x^{4}

代回

= - \frac{1}{12} \cdot \frac{3}{4} (8 - 3 x^{4})^{\frac{4}{3}}

化简

- \frac{1}{12} \cdot \frac{3}{4} (8 - 3 x^{4})^{\frac{4}{3}} : - \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}}

= - \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}}

解答补常数

= - \frac{1}{16} (8 - 3 x^{4})^{\frac{4}{3}} + C

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