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

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

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

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

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

Solution steps

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

Simplify

ln (x^{2} - 1) : \frac{1}{2} ln (x^{2} - 1)

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

Take the constant out:

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

= \frac{1}{2} \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

= \frac{1}{2} (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

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

Simplify

\frac{1}{2} (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x)) : \frac{1}{2} x ln (x^{2} - 1) + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x

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

Add a constant to the solution

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

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