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

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

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integral of x/(x^2-1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 1

= \frac{1}{2} ln x^{2} - 1

Add a constant to the solution

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

x \to 3 lim (x + 1 - 2)

\int csc (u) d u

\frac{d}{d x} (\frac{- y}{x ^{2} + y ^{2}})

\int \frac{( 1 - 3 y )}{2 y - 3 y ^{2}} d y

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