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

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

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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

\frac{x ^{3}}{x}

x (x^{2} + 1)^{3}

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

x^{- 2} e^{x}

\frac{y}{2}

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