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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int e^{- t x} d x

\int \frac{2}{25 - 2 x ^{2}} d x

6 x + \frac{4}{x}

n = 1 \sum \infty \frac{1}{n ( ln ( n ) )}

x cos (x) sin (x)

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