integral of x^4 1/(sqrt(1-x^2))

解答

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

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

求解步骤

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

化简

x^{4} \frac{1}{1 - x ^{2}} : \frac{x ^{4}}{1 - x ^{2}}

= \int \frac{x ^{4}}{1 - x ^{2}} d x

使用三角换元法

= \int sin^{4} (u) d u

使用降次积分法

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u

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

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

u = arcsin (x)

代回

= - \frac{cos ( arcsin ( x ) ) sin ^{3} ( arcsin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

化简

- \frac{cos ( arcsin ( x ) ) sin ^{3} ( arcsin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) : - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

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

解答补常数

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

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