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)))

integral

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

- 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

e^{x} \cdot (\frac{1}{x})

\frac{1}{( 3 t ^{4} + 4 t ^{2} + 1 )}

\frac{( x - 2 ) ^{3}}{3 x ^{2}}

\frac{t ^{3}}{( 1 + t ^{6} )}

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

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