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受欢迎的微积分 >

integral of 1/(sqrt(8-4x-4x^2))

解答

\int \frac{1}{8 - 4 x - 4 x ^{2}} d x

解答

\frac{1}{2} arcsin (\frac{2}{3} x + \frac{1}{3}) + C

求解步骤

\int \frac{1}{8 - 4 x - 4 x ^{2}} d x

对

8 - 4 x - 4 x^{2}

配方:

- 4 (x + \frac{1}{2})^{2} + 9

= \int \frac{1}{- 4 ( x + \frac{1}{2} ) ^{2} + 9} d x

使用换元积分法

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

使用三角换元法

= \int \frac{1}{2} d v

提出常数:

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

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

常数积分:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

代回

= \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{3} (x + \frac{1}{2}))

化简

\frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{3} (x + \frac{1}{2})) : \frac{1}{2} arcsin (\frac{2}{3} x + \frac{1}{3})

= \frac{1}{2} arcsin (\frac{2}{3} x + \frac{1}{3})

解答补常数

= \frac{1}{2} arcsin (\frac{2}{3} x + \frac{1}{3}) + C

作图

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