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

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

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

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

ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

\int \frac{x ^{2} + 58}{x ^{2} + 64} d x

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