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

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

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integral of 1/(u(1-u))

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

ln ∣ u ∣ - ln ∣ u - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{u} - \frac{1}{u - 1} d u

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}{u} d u - \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= ln ∣ u ∣ - ln ∣ u - 1 ∣

Add a constant to the solution

= ln ∣ u ∣ - ln ∣ u - 1 ∣ + C

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\frac{\partial}{\partial y} (x 1 + y^{2})

\frac{\partial}{\partial y} (sin (x) cos (7 y))

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