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))

integral

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

- 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{1}{sin ( 2 x )}

\frac{e ^{- z}}{( x ^{2} + y ^{2} )}

\frac{1}{2 x - 1}

1 + t^{2}

\frac{( x ^{2} + 2 x + 1 )}{x ^{2}}

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