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受欢迎的微积分 >

integral of (2u)/(sqrt(-4u^2+9))

解答

\int \frac{2 u}{- 4 u ^{2} + 9} d u

解答

- \frac{1}{2} - 4 u^{2} + 9 + C

求解步骤

\int \frac{2 u}{- 4 u ^{2} + 9} d u

提出常数:

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

= 2 \cdot \int \frac{u}{- 4 u ^{2} + 9} d u

使用换元积分法

= 2 \cdot \int - \frac{1}{8 v} d v

提出常数:

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

= 2 (- \frac{1}{8} \cdot \int \frac{1}{v} d v)

使用幂法则

= 2 (- \frac{1}{8} \cdot 2 v^{\frac{1}{2}})

v = - 4 u^{2} + 9

代回

= 2 (- \frac{1}{8} \cdot 2 (- 4 u^{2} + 9)^{\frac{1}{2}})

化简

2 (- \frac{1}{8} \cdot 2 (- 4 u^{2} + 9)^{\frac{1}{2}}) : - \frac{1}{2} - 4 u^{2} + 9

= - \frac{1}{2} - 4 u^{2} + 9

解答补常数

= - \frac{1}{2} - 4 u^{2} + 9 + C

作图

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