integral of x/(\sqrt[3]{x-6)}

解答

\int \frac{x}{3 x - 6} d x

\frac{3}{5} (x - 6)^{\frac{5}{3}} + 9 (x - 6)^{\frac{2}{3}} + C

求解步骤

\int \frac{x}{3 x - 6} d x

使用换元积分法

= \int \frac{u + 6}{3 u} d u

乘开

\frac{u + 6}{3 u} : u^{\frac{2}{3}} + \frac{6}{3 u}

= \int u^{\frac{2}{3}} + \frac{6}{3 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

= \int u^{\frac{2}{3}} d u + \int \frac{6}{3 u} d u

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

\int \frac{6}{3 u} d u = 9 u^{\frac{2}{3}}

= \frac{3}{5} u^{\frac{5}{3}} + 9 u^{\frac{2}{3}}

u = x - 6

代回

= \frac{3}{5} (x - 6)^{\frac{5}{3}} + 9 (x - 6)^{\frac{2}{3}}

解答补常数

= \frac{3}{5} (x - 6)^{\frac{5}{3}} + 9 (x - 6)^{\frac{2}{3}} + C

\frac{\partial}{\partial x} (y^{2} - 4 x)

\frac{d}{d x} (\frac{x ^{3}}{1 + x ^{2}})

n = 0 \sum \infty \frac{n + 1}{2 n - 3}

\int \frac{2 x - 2}{1 + ( x - 1 ) ^{2}} d x

\frac{d x}{d t} = 7 (x^{2} + 1), x (\frac{π}{4}) = 1