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

\frac{d}{d x} (2 (x^{2} + 5 x - 2)^{- \frac{5}{7}})

\frac{d ^{2} y}{d t ^{2}} + 9 y = 0

\frac{d}{d x} (csc^{2} (x - 2))

\int_{0}^{1} x (1 - x)^{5} d x

\frac{\partial}{\partial x} (x^{0.4} y^{0.6})

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