What is the integral of-8ln(x^2-1) ?

The integral of-8ln(x^2-1) is -8(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C

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integral of-8ln(x^2-1)

\int - 8 ln (x^{2} - 1) d x

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

Solution steps

\int - 8 ln (x^{2} - 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 8 \cdot \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

= - 8 (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

= - 8 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= - 8 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= - 8 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

\int_{- 2}^{3} (x^{2} - 12) - (x - 6) d x

x \to 1 lim (x - 1)

x \to 1 lim (3 e^{x - 1})

y = \frac{1}{x - 1}

x \to - 1 lim (x ∣ x - 4 ∣)

What is the integral of-8ln(x^2-1) ?
The integral of-8ln(x^2-1) is -8(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C