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受欢迎的微积分 >

integral of (9x)/(\sqrt[4]{x+2)}

解答

\int \frac{9 x}{4 x + 2} d x

解答

9 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}) + C

求解步骤

\int \frac{9 x}{4 x + 2} d x

提出常数:

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

= 9 \cdot \int \frac{x}{4 x + 2} d x

使用换元积分法

= 9 \cdot \int \frac{u - 2}{4 u} d u

乘开

\frac{u - 2}{4 u} : u^{\frac{3}{4}} - \frac{2}{4 u}

= 9 \cdot \int u^{\frac{3}{4}} - \frac{2}{4 u} d u

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

= 9 (\int u^{\frac{3}{4}} d u - \int \frac{2}{4 u} d u)

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

\int \frac{2}{4 u} d u = \frac{8}{3} u^{\frac{3}{4}}

= 9 (\frac{4}{7} u^{\frac{7}{4}} - \frac{8}{3} u^{\frac{3}{4}})

u = x + 2

代回

= 9 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}})

解答补常数

= 9 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}) + C

作图

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