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

y^{^{'}} + \frac{x}{x ^{2} + 1} y = - 4 x

y = (\frac{( x + 2 )}{( x - 3 )})^{2}, at x = 4

\frac{d}{d x} (a (e^{x} + e^{x} x))

h \to 0 lim (h sin (\frac{1}{h}))

\int \frac{1}{6} x^{2} d x

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