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

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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

\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

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