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

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

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

integral

\frac{1}{( x - x ^{2} )}

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

in t e g r a l sin (2 x^{2})

in t e g r a l \frac{( x + 1 ) ^{3}}{x ^{3}}

in t e g r a l y = x^{2} + x - 12

in t e g r a l \frac{x + 1}{( x ^{2} + 4 x + 8 )}

in t e g r a l \frac{u}{( 1 + u )}

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