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受欢迎的微积分 >

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

解答

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

解答

x + 1 x - 1 + ln \frac{x + 1}{x - 1} + 1 - ln \frac{x + 1}{x - 1} - 1 + C

求解步骤

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

使用换元积分法

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

\frac{u + 2}{u} = (\frac{u}{u} + \frac{2}{u})

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

\frac{u}{u} = 1

= \int 1 + \frac{2}{u} d u

使用换元积分法

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

提出常数:

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

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

使用分布积分法

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

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

= - 2 (- \frac{v + 1}{v} - (\frac{1}{2} ln v + 1 + 1 - \frac{1}{2} ln v + 1 - 1))

代回

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

化简

- 2 - \frac{\frac{2}{x - 1} + 1}{\frac{2}{x - 1}} - (\frac{1}{2} ln \frac{2}{x - 1} + 1 + 1 - \frac{1}{2} ln \frac{2}{x - 1} + 1 - 1) : x + 1 x - 1 + ln \frac{x + 1}{x - 1} + 1 - ln \frac{x + 1}{x - 1} - 1

= x + 1 x - 1 + ln \frac{x + 1}{x - 1} + 1 - ln \frac{x + 1}{x - 1} - 1

解答补常数

= x + 1 x - 1 + ln \frac{x + 1}{x - 1} + 1 - ln \frac{x + 1}{x - 1} - 1 + C

作图

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