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

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

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integral of 1ln(x^2-1)

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

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

Solution steps

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

Simplify

= \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

e^{x}, at x = 5

\frac{d}{d x} (x^{5} sin (x) tan (x))

x \to - 5 - lim (\frac{8}{( - 5 ) + 5})

\frac{d}{d x} ((3 x - 1)^{5})

\int \frac{1}{x ^{2} y} d y

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