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受欢迎的微积分 >

integral of sqrt((1+2x)/(1-2x))

解答

\int \frac{1 + 2 x}{1 - 2 x} d x

解答

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

求解步骤

\int \frac{1 + 2 x}{1 - 2 x} d x

使用换元积分法

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

使用换元积分法

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

提出常数:

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

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

使用三角换元法

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

使用三角恒等式改写

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

代回

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

化简

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

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

解答补常数

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

作图

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\int r ln (\frac{2 r}{g}) d r

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