What is the integral of 7ln(x^2-1) ?

The integral of 7ln(x^2-1) is 7(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C

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integral of 7ln(x^2-1)

\int 7 ln (x^{2} - 1) d x

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

Solution steps

\int 7 ln (x^{2} - 1) d x

Take the constant out:

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

= 7 \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

= 7 (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

= 7 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= 7 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= 7 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

\frac{\partial}{\partial z} (x e^{y} + y e^{z} + z e^{x})

\int (1 - x^{2})^{0.5} d x

\frac{2}{4 - x}, at x = 1

\int sin^{2} (4 x) cos^{2} (x) 4 x d x

\frac{d y}{d x} = 5 x e^{y}, y (0) = 0

What is the integral of 7ln(x^2-1) ?
The integral of 7ln(x^2-1) is 7(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C