What is the integral of-6ln(x^2-1) ?

The integral of-6ln(x^2-1) is -6(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C

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integral of-6ln(x^2-1)

\int - 6 ln (x^{2} - 1) d x

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

Solution steps

\int - 6 ln (x^{2} - 1) d x

Take the constant out:

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

= - 6 \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

= - 6 (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

= - 6 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= - 6 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= - 6 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

\int (x cos (4 x)) d x

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

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

\frac{d}{d x} (x e^{- x} - e^{- x})

n = 1 \sum \infty n 3^{- n}

What is the integral of-6ln(x^2-1) ?
The integral of-6ln(x^2-1) is -6(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C