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受欢迎的微积分 >

integral of sqrt((3-2x)/(3+2x))

解答

\int \frac{3 - 2 x}{3 + 2 x} d x

解答

3 (arcsin (\frac{1}{6} 3 + 2 x) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} 3 + 2 x))) + C

求解步骤

\int \frac{3 - 2 x}{3 + 2 x} d x

使用换元积分法

= \int \frac{- u + 6}{2 u} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{- u + 6}{u} d u

使用换元积分法

= \frac{1}{2} \cdot \int 2 - v^{2} + 6 d v

提出常数:

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

= \frac{1}{2} \cdot 2 \cdot \int - v^{2} + 6 d v

使用三角换元法

= \frac{1}{2} \cdot 2 \cdot \int 6 cos^{2} (w) d w

提出常数:

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

= \frac{1}{2} \cdot 2 \cdot 6 \cdot \int cos^{2} (w) d w

使用三角恒等式改写

= \frac{1}{2} \cdot 2 \cdot 6 \cdot \int \frac{1 + cos ( 2 w )}{2} d w

提出常数:

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

= \frac{1}{2} \cdot 2 \cdot 6 \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot 2 \cdot 6 \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= \frac{1}{2} \cdot 2 \cdot 6 \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

代回

= \frac{1}{2} \cdot 2 \cdot 6 \cdot \frac{1}{2} (arcsin (\frac{1}{6} 3 + 2 x) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} 3 + 2 x)))

化简

\frac{1}{2} \cdot 2 \cdot 6 \cdot \frac{1}{2} (arcsin (\frac{1}{6} 3 + 2 x) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} 3 + 2 x))) : 3 (arcsin (\frac{1}{6} 3 + 2 x) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} 3 + 2 x)))

= 3 (arcsin (\frac{1}{6} 3 + 2 x) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} 3 + 2 x)))

解答补常数

= 3 (arcsin (\frac{1}{6} 3 + 2 x) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} 3 + 2 x))) + C

作图

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