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

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

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

\int \frac{1}{x ( x - 1 )} d x

- ln ∣ x ∣ + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{1}{x ( x - 1 )} d x

Take the partial fraction of

\frac{1}{x ( x - 1 )} : - \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

\int_{6}^{7} x^{π - 1} d x

x \to 1 lim (\frac{1}{2 x - 1})

\frac{d}{d x} ((2 x^{2} - 1)^{6})

\int e^{- 2 x} \frac{d}{1} (y) - 2 e^{- 2 x} y d x

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{3}})

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