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

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

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integral of ln(x^2-1)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{2} ln (3 y - 5))

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

n = 2 \sum \infty \frac{1}{n ^{2} - 1}

\frac{\partial}{\partial z} (ln (x + 2 y + 3 z))

x \to 3 lim (\frac{x ^{2} + x}{x - 3})

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