integral of \sqrt[3]{1-2x}

解答

\int 3 1 - 2 x d x

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

求解步骤

\int 3 1 - 2 x d x

使用换元积分法

= \int - \frac{3 u}{2} d u

提出常数:

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

= - \frac{1}{2} \cdot \int 3 u d u

使用幂法则

= - \frac{1}{2} \cdot \frac{3}{4} u^{\frac{4}{3}}

u = 1 - 2 x

代回

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

化简

- \frac{1}{2} \cdot \frac{3}{4} (1 - 2 x)^{\frac{4}{3}} : - \frac{3}{8} (1 - 2 x)^{\frac{4}{3}}

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

解答补常数

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

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