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

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

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integral of (x-1)/(x^2-1)

\int \frac{x - 1}{x ^{2} - 1} d x

ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{x - 1}{x ^{2} - 1} d x

Simplify

\frac{x - 1}{x ^{2} - 1} : \frac{1}{x + 1}

= \int \frac{1}{x + 1} d x

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x + 1

= ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ + C

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