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

解答

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

\frac{1}{8} (\frac{3}{7} (1 + 2 x^{2})^{\frac{7}{3}} - \frac{3}{4} (1 + 2 x^{2})^{\frac{4}{3}}) + C

求解步骤

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

使用换元积分法

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

乘开

3 u (u - 1) : u^{\frac{4}{3}} - 3 u

= \frac{1}{8} \cdot \int u^{\frac{4}{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{4}{3}} d u - \int 3 u d u)

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

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

= \frac{1}{8} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

u = 1 + 2 x^{2}

代回

= \frac{1}{8} (\frac{3}{7} (1 + 2 x^{2})^{\frac{7}{3}} - \frac{3}{4} (1 + 2 x^{2})^{\frac{4}{3}})

解答补常数

= \frac{1}{8} (\frac{3}{7} (1 + 2 x^{2})^{\frac{7}{3}} - \frac{3}{4} (1 + 2 x^{2})^{\frac{4}{3}}) + C

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