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)

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

4 s

\frac{\partial ^{2}}{\partial x \partial y} (x y e^{- 6 y})

n = 1 \sum \infty \frac{2}{n ^{2} + n}

x \to 0 lim (\frac{( x - 1 ) ^{3} + 1}{h ( x )})

x \to + 1 lim (x sin (\frac{1}{x}))

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