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

解答

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

- 2 ln ∣ x + 1 ∣ + x + C

求解步骤

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

\frac{x - 1}{x + 1} = - \frac{2}{x + 1} + 1

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

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

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

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

\int 1 d x = x

= - 2 ln ∣ x + 1 ∣ + x

解答补常数

= - 2 ln ∣ x + 1 ∣ + x + C

\int \frac{( e ^{x} + 1 )}{( e ^{x} - 1 )} d x

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