What is the integral of 3ln(x^2-1) ?

The integral of 3ln(x^2-1) is 3(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C

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integral of 3ln(x^2-1)

\int 3 ln (x^{2} - 1) d x

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

Solution steps

\int 3 ln (x^{2} - 1) d x

Take the constant out:

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

= 3 \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

= 3 (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

= 3 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= 3 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= 3 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

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

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = 1

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

f (x) = 4 sec (x), at \frac{π}{3}

\int (3 x^{2} (2 x^{3} - 1)) d x

What is the integral of 3ln(x^2-1) ?
The integral of 3ln(x^2-1) is 3(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C