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

(2 + x) y^{^{'}} = 8 y

x \to 0 lim (e^{x} - e^{- x})

\frac{d y}{d x} = 9 e^{x - y}

f (x) = 3 x^{3} (x^{2} - 3 x + 5)

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