What is the integral of ln((x+1)/(x-1)) ?

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

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integral of ln((x+1)/(x-1))

\int ln (\frac{x + 1}{x - 1}) d x

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

Solution steps

\int ln (\frac{x + 1}{x - 1}) d x

Apply Integration By Parts

: x ln (\frac{x + 1}{x - 1}) - \int - \frac{2 x}{( x + 1 ) ( x - 1 )} d x

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

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

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

Simplify

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

Add a constant to the solution

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

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