integral of (x^2)/(\sqrt[3]{1+2x)}

解答

\int \frac{x ^{2}}{3 1 + 2 x} d x

\frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\frac{2}{3}}) + C

求解步骤

\int \frac{x ^{2}}{3 1 + 2 x} d x

使用换元积分法

= \int \frac{( u - 1 ) ^{2}}{8 3 u} d u

提出常数:

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

= \frac{1}{8} \cdot \int \frac{( u - 1 ) ^{2}}{3 u} d u

乘开

\frac{( u - 1 ) ^{2}}{3 u} : u^{\frac{5}{3}} - 2 u^{\frac{2}{3}} + \frac{1}{3 u}

= \frac{1}{8} \cdot \int u^{\frac{5}{3}} - 2 u^{\frac{2}{3}} + \frac{1}{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

= \frac{1}{8} (\int u^{\frac{5}{3}} d u - \int 2 u^{\frac{2}{3}} d u + \int \frac{1}{3 u} d u)

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 2 u^{\frac{2}{3}} d u = \frac{6}{5} u^{\frac{5}{3}}

\int \frac{1}{3 u} d u = \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{8} (\frac{3}{8} u^{\frac{8}{3}} - \frac{6}{5} u^{\frac{5}{3}} + \frac{3}{2} u^{\frac{2}{3}})

u = 1 + 2 x

代回

= \frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\frac{2}{3}})

解答补常数

= \frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\frac{2}{3}}) + C

\frac{d}{d x} ((x - 2)^{2})

2 (x y - y) d x - x d y = 0

\int \frac{1}{x ^{\frac{1}{2}}} d x

\int x \cdot ln (5 + x) d x

\int \frac{π}{2} - 5 x^{- 0.5} d x