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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C

Solution steps

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

\frac{x ^{2} + 1}{x ^{2} - 1} = \frac{2}{x ^{2} - 1} + 1

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

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

\int 1 d x = x

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x

Add a constant to the solution

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C

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