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

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

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

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

ln x^{2} - x + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= ln x^{2} - x + x

Add a constant to the solution

= ln x^{2} - x + x + C

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