integral of (x^2(3-x))/(sqrt(x^4-4x^3))

解答

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

- \frac{1}{2} x^{4} - 4 x^{3} + C

求解步骤

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

使用换元积分法

= \int - \frac{1}{4 u} d u

提出常数:

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

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

使用幂法则

= - \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

u = x^{4} - 4 x^{3}

代回

= - \frac{1}{4} \cdot 2 (x^{4} - 4 x^{3})^{\frac{1}{2}}

化简

- \frac{1}{4} \cdot 2 (x^{4} - 4 x^{3})^{\frac{1}{2}} : - \frac{1}{2} x^{4} - 4 x^{3}

= - \frac{1}{2} x^{4} - 4 x^{3}

解答补常数

= - \frac{1}{2} x^{4} - 4 x^{3} + C

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