What is the integral of 1/(2x(1-x)) ?

The integral of 1/(2x(1-x)) is 1/2 (ln|x|-ln|x-1|)+C

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integral of 1/(2x(1-x))

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

\frac{1}{2} (ln ∣ x ∣ - ln ∣ x - 1 ∣) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \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}

= \frac{1}{2} \cdot \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

= \frac{1}{2} (\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 ∣

= \frac{1}{2} (ln ∣ x ∣ - ln ∣ x - 1 ∣)

Add a constant to the solution

= \frac{1}{2} (ln ∣ x ∣ - ln ∣ x - 1 ∣) + C

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