What is the solution for 1/(p-p^2)dp=dt ?

The solution for 1/(p-p^2)dp=dt is t=ln(p)-ln(p-1)+c_{1}

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1/(p-p^2)dp=dt

\frac{1}{p - p ^{2}} d p = d t

t = ln (p) - ln (p - 1) + c_{1}

Solution steps

\frac{1}{p - p ^{2}} d p = d t

Let

t

be the dependent variable. Divide by

d p

\frac{1}{p - p ^{2}} = \frac{d t}{d p}

Substitute

\frac{d t}{d p}

with

t^{'} (p)

\frac{1}{p - p ^{2}} = t^{^{'}} (p)

Isolate

t^{'} (p) : t^{'} (p) = \frac{1}{p - p ^{2}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{p - p ^{2}} d p

\int \frac{1}{p - p ^{2}} d p = ln (p) - ln (p - 1) + c_{1}

t = ln (p) - ln (p - 1) + c_{1}

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What is the solution for 1/(p-p^2)dp=dt ?
The solution for 1/(p-p^2)dp=dt is t=ln(p)-ln(p-1)+c_{1}