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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

\int \frac{3}{2 3 ( 4 x - 1 ) ^{2}} d x

\int \frac{x - 2}{x ^{2} - 4 x + 7} d x

\int - \frac{4 cos ( x )}{3 - 10 sin ( x )} d x

\int \frac{1}{x ^{2} + 4 x - 2} d x

\int 6 x^{2} - 6 d x

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