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

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

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integral of 1/(t(1-t))

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

ln ∣ t ∣ - ln ∣ t - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

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

= ln ∣ t ∣ - ln ∣ t - 1 ∣

Add a constant to the solution

= ln ∣ t ∣ - ln ∣ t - 1 ∣ + C

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