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

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

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integral of 1/(x-x^2)

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

ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

\frac{d}{d x} (π \cdot x^{2})

y = \frac{( 3 x ^{5} + 9 )}{x ^{3}}

\frac{d}{d s} (ln (s^{2} + 9))

4 t^{- \frac{1}{8}}

r (x) = \frac{3}{x ^{4} + 2}

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