integral of 1/(\sqrt[4]{6-9x)}

解答

\int \frac{1}{4 6 - 9 x} d x

- \frac{4}{27} (6 - 9 x)^{\frac{3}{4}} + C

求解步骤

\int \frac{1}{4 6 - 9 x} d x

使用换元积分法

= \int - \frac{1}{9 4 u} d u

提出常数:

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

= - \frac{1}{9} \cdot \int \frac{1}{4 u} d u

使用幂法则

= - \frac{1}{9} \cdot \frac{4}{3} u^{\frac{3}{4}}

u = 6 - 9 x

代回

= - \frac{1}{9} \cdot \frac{4}{3} (6 - 9 x)^{\frac{3}{4}}

化简

- \frac{1}{9} \cdot \frac{4}{3} (6 - 9 x)^{\frac{3}{4}} : - \frac{4}{27} (6 - 9 x)^{\frac{3}{4}}

= - \frac{4}{27} (6 - 9 x)^{\frac{3}{4}}

解答补常数

= - \frac{4}{27} (6 - 9 x)^{\frac{3}{4}} + C

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