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

\frac{\partial}{\partial x} (y cos (x - y))

64 y^{^{''}} + 80 y^{^{'}} + 25 y = 0, y (0) = a, y^{^{'}} (0) = - 5

f (x) = ln (x^{2} - 9 x + 1), at x = 9

\int e^{- x - y} d x

\int 7 arcsin (x) d x

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