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

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

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

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

x - 1 + ln ∣ x - 1 ∣ + C

Solution steps

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

Simplify

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

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

Apply u-substitution

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

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u + ln ∣ u ∣

Substitute back

u = x - 1

= x - 1 + ln ∣ x - 1 ∣

Add a constant to the solution

= x - 1 + ln ∣ x - 1 ∣ + C

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