In this question, we have to find the** Integration** of the given function $ \dfrac{dP}{dt}= \left[P â€“ P^{2} \right] $ by rearranging the equation.

The basic concept behind this question is the knowledge of **derivatives, integration,** and the **rules** such as the **product and quotient rules** of **integration**.

**Expert Answer**

Given function:

\[\dfrac{dP}{dt}= \left[P â€“ P^{2} \right] \]

First, we will** rearrange** the **given equation** with $P $ on one side of the equation and $t $ on the other. For this, we have the following equation:

\[dP = \left[P â€“ P^{2} \right] {dt} \]

\[\dfrac{1 }{\left[P â€“ P^{2} \right]} dP = dt \]

\[ dt =\dfrac{1 }{\left[P â€“ P^{2} \right]} dP \]

Take **Integration** on both sides of the equation. We get:

\[ \int dt = \int \dfrac{1 }{P â€“ P^{2}} dP \]

Taking $P $ common on the **right-hand side,** we will have the equation:

\[ \int dt = \int \dfrac{1 }{P (1 â€“ P)} dP\]

As we can write $ 1 = ( 1-P ) + P $ in the **above equation**, putting it in the question we have the following equation:

\[ \int dt = \int \dfrac{(1-P) + P }{P (1 â€“ P)} dP \]

\[ \int dt = \int \dfrac{(1-P) }{P (1 â€“ P)} dP + \int \dfrac{P }{P (1 â€“ P)} Â dP \]

Canceling $ 1-P$ from **the denominator **and** numerator** of the equation:

\[ \int dt = \int \dfrac{1 }{P } dP + \int \dfrac{P }{P (1 â€“ P)} Â dP\]

Canceling $ P$ from **the denominator **and **numerator** of the equation:

\[ \int dt = \int \dfrac{1 }{P } dP + \int \dfrac{1 }{ (1 â€“ P)} dP\]

Solving the** above equation** now:

\[ t + c_1 = \ln{\left|P \right|\ -\ }\ln{\left|1-P\right|\ } \]

\[ t + c_1 =\ln{\left|\ \frac{ P }{ 1 – P }\ \right|} \]

\[ e^{ t + c_{1} } =e^{\ln{\left|\ \dfrac{ P }{ 1 – P }\ \right|}} \]

We know that $ e^{\ln{x} } = x $ so we have the above **equation** as:

\[ e^{ t} e^{ c_1 } = \left| \dfrac { P }{ 1 – P } Â \right| \]

\[ \left| \dfrac { P }{ 1-P } Â \right| = e^{ t} e^{ c_1 } \]

\[ Â \dfrac { P }{ 1-P } Â = \pm e^{ t} e^{ c_1 } \]

Let us suppose that **another constant** $c $ is **introduced **in the **equation** which is $ \pm e^{ c_1 } Â = c $. Now the** equation** becomes:

\[ Â \dfrac { P }{ 1-P } Â = ce^{ t} Â \]

**Multiplying** by $ 1-P $ on both the side of the equation:

\[ P=c e^t (1-P) \]

\[ P = ce^t- ce^{t}P\]

\[P+ ce^{t}P = ce^t\]

\[P(1+ ce^{t}) = ce^t\]

\[P= \dfrac{ce^t}{(1+ ce^{t})}\]

## Numerical Result

\[P= \dfrac{ce^t}{(1+ ce^{t})}\]

## Example

**Integrate** the equation:

\[\int dt= \int \dfrac{1}{x } dx \]

Solving the **above equation** now:

\[t+c_1 = \ln{\left|x \right|}\]

\[e^{t+ c_1}=e^{\ln{x}}\]

We know that $ e^{\ln{x}} = x $ so we have the above** equation** as:

\[e^{t} e^{ c_1}=x\]