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

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

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integral of 1/(P-P^2)

\int \frac{1}{P - P ^{2}} d P

ln ∣ P ∣ - ln ∣ P - 1 ∣ + C

Solution steps

\int \frac{1}{P - P ^{2}} d P

Take the partial fraction of

\frac{1}{P - P ^{2}} : \frac{1}{P} - \frac{1}{P - 1}

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

\int \frac{1}{P} d P = ln ∣ P ∣

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

= ln ∣ P ∣ - ln ∣ P - 1 ∣

Add a constant to the solution

= ln ∣ P ∣ - ln ∣ P - 1 ∣ + C

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