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

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

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

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

x + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

= x + ln ∣ x ∣

Add a constant to the solution

= x + ln ∣ x ∣ + C

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