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

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

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

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

ln ∣ x ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ( 1 - 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

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

\frac{d}{d x} ((3 x^{2} - x)^{7})

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

\frac{d}{d x} (10 sec (2 x) tan (2 x))

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

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