What is the integral of 1/(p(5-p)) ?

The integral of 1/(p(5-p)) is 1/5 ln|p|-1/5 ln|p-5|+C

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integral of 1/(p(5-p))

\int \frac{1}{p ( 5 - p )} d p

\frac{1}{5} ln ∣ p ∣ - \frac{1}{5} ln ∣ p - 5 ∣ + C

Solution steps

\int \frac{1}{p ( 5 - p )} d p

Take the partial fraction of

\frac{1}{p ( 5 - p )} : \frac{1}{5 p} - \frac{1}{5 ( p - 5 )}

= \int \frac{1}{5 p} - \frac{1}{5 ( p - 5 )} d p

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}{5 p} d p - \int \frac{1}{5 ( p - 5 )} d p

\int \frac{1}{5 p} d p = \frac{1}{5} ln ∣ p ∣

\int \frac{1}{5 ( p - 5 )} d p = \frac{1}{5} ln ∣ p - 5 ∣

= \frac{1}{5} ln ∣ p ∣ - \frac{1}{5} ln ∣ p - 5 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ p ∣ - \frac{1}{5} ln ∣ p - 5 ∣ + C

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What is the integral of 1/(p(5-p)) ?
The integral of 1/(p(5-p)) is 1/5 ln|p|-1/5 ln|p-5|+C