integral of x/(sqrt(25-4x^2))

解答

\int \frac{x}{25 - 4 x ^{2}} d x

- \frac{1}{4} 25 - 4 x^{2} + C

求解步骤

\int \frac{x}{25 - 4 x ^{2}} d x

使用换元积分法

= \int - \frac{1}{8 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{1}{u} d u

使用幂法则

= - \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

u = 25 - 4 x^{2}

代回

= - \frac{1}{8} \cdot 2 (25 - 4 x^{2})^{\frac{1}{2}}

化简

- \frac{1}{8} \cdot 2 (25 - 4 x^{2})^{\frac{1}{2}} : - \frac{1}{4} 25 - 4 x^{2}

= - \frac{1}{4} 25 - 4 x^{2}

解答补常数

= - \frac{1}{4} 25 - 4 x^{2} + C

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