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

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

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

- \frac{ln ( x ^{2} - 1 )}{x} - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

= - \frac{ln ( x ^{2} - 1 )}{x} - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

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

\int \frac{5}{3 4 x ^{5}} d x

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

\int 27 x e^{3 x} d x

\int (e^{x} + 4^{x}) d x

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