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

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

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

integral

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

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

in t e g r a l sin (2 x - y)

in t e g r a l \frac{1}{2 sin ( 2 x )}

in t e g r a l \frac{x ^{3}}{( x ^{2} + x + 2 )}

in t e g r a l ln (x - 1)

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