What is the integral of 10ln(x^2-1) ?

The integral of 10ln(x^2-1) is 10(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C

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integral of 10ln(x^2-1)

\int 10 ln (x^{2} - 1) d x

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

Solution steps

\int 10 ln (x^{2} - 1) d x

Take the constant out:

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

= 10 \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

= 10 (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

= 10 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= 10 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= 10 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

\frac{2 x - 6}{x + 1}, at x = 0

\frac{t}{t ^{2} + 36}

e^{x} \frac{d y}{d x} = 2

y^{^{''}} + y^{^{'}} + 4.25 y = 221 cos (4.5 t)

\int_{0}^{4} \frac{1}{4 - x} d x

What is the integral of 10ln(x^2-1) ?
The integral of 10ln(x^2-1) is 10(xln(x^2-1)+ln|x+1|-ln|x-1|-2x)+C