What is the integral of ((x-1)/(x+1)) ?

The integral of ((x-1)/(x+1)) is -2ln|x+1|+x+C

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

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

- 2 ln ∣ x + 1 ∣ + x + C

Solution steps

\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

Apply the Sum Rule:

\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

Add a constant to the solution

= - 2 ln ∣ x + 1 ∣ + x + C

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What is the integral of ((x-1)/(x+1)) ?
The integral of ((x-1)/(x+1)) is -2ln|x+1|+x+C