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

解答

\int \frac{x ^{3}}{3 3 - 2 x ^{2}} d x

\frac{1}{8} (\frac{3}{5} (3 - 2 x^{2})^{\frac{5}{3}} - \frac{9}{2} (3 - 2 x^{2})^{\frac{2}{3}}) + C

求解步骤

\int \frac{x ^{3}}{3 3 - 2 x ^{2}} d x

使用换元积分法

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

乘开

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

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

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

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

= \frac{1}{8} (\frac{3}{5} u^{\frac{5}{3}} - \frac{9}{2} u^{\frac{2}{3}})

u = 3 - 2 x^{2}

代回

= \frac{1}{8} (\frac{3}{5} (3 - 2 x^{2})^{\frac{5}{3}} - \frac{9}{2} (3 - 2 x^{2})^{\frac{2}{3}})

解答补常数

= \frac{1}{8} (\frac{3}{5} (3 - 2 x^{2})^{\frac{5}{3}} - \frac{9}{2} (3 - 2 x^{2})^{\frac{2}{3}}) + C

x \to - 2 lim (\frac{3 x - 9}{x + 1})

n = 2 \sum \infty \frac{1}{( - 5 ) ^{n}}

n \to + 0 - lim (5 + \frac{4}{n})

\int 3 \cdot cos (3 x) d x

\int \frac{x ^{15}}{x ^{16} + 1} d x