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

n = 0 \sum \infty 2^{- n} + 3^{- n}

\frac{d}{d x} (\frac{1}{( x + 1 )})

\frac{d}{d x} (arccsc (- x^{2}))

\int (5 x^{2}) d x

\int \frac{tan ^{2} ( x )}{2 sec ( x )} d x

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