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) with respect to x

\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} (\frac{x}{a})

- \frac{d}{d x} (2 e^{- 2 x})

- \frac{d}{d} y y^{2}

\frac{- 1}{( e ^{x} )}

- \frac{d}{d x} y = 4 x

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