What is the integral of (2x+1)/(x-1) ?

The integral of (2x+1)/(x-1) is 2x-2+3ln|x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x+1)/(x-1)

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

2 x - 2 + 3 ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= 2 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣

Simplify

2 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣ : 2 x - 2 + 3 ln ∣ x - 1 ∣

= 2 x - 2 + 3 ln ∣ x - 1 ∣

Add a constant to the solution

= 2 x - 2 + 3 ln ∣ x - 1 ∣ + C

What is the integral of (2x+1)/(x-1) ?
The integral of (2x+1)/(x-1) is 2x-2+3ln|x-1|+C