integral of x^3(1-4x^2)^{3/2}

解答

\int x^{3} (1 - 4 x^{2})^{\frac{3}{2}} d x

- \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}} + C

求解步骤

\int x^{3} (1 - 4 x^{2})^{\frac{3}{2}} d x

使用换元积分法

= \int \frac{u ^{\frac{3}{2}} ( u - 1 )}{32} d u

提出常数:

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

= \frac{1}{32} \cdot \int u^{\frac{3}{2}} (u - 1) d u

使用分布积分法

= \frac{1}{32} (\frac{2}{5} u^{\frac{5}{2}} (u - 1) - \int \frac{2}{5} u^{\frac{5}{2}} d u)

\int \frac{2}{5} u^{\frac{5}{2}} d u = \frac{4}{35} u^{\frac{7}{2}}

= \frac{1}{32} (\frac{2}{5} u^{\frac{5}{2}} (u - 1) - \frac{4}{35} u^{\frac{7}{2}})

u = 1 - 4 x^{2}

代回

= \frac{1}{32} (\frac{2}{5} (1 - 4 x^{2})^{\frac{5}{2}} (1 - 4 x^{2} - 1) - \frac{4}{35} (1 - 4 x^{2})^{\frac{7}{2}})

化简

\frac{1}{32} (\frac{2}{5} (1 - 4 x^{2})^{\frac{5}{2}} (1 - 4 x^{2} - 1) - \frac{4}{35} (1 - 4 x^{2})^{\frac{7}{2}}) : - \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}}

= - \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}}

解答补常数

= - \frac{1}{280} (10 x^{2} + 1) (- 4 x^{2} + 1)^{\frac{5}{2}} + C

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