Hint request for $x' = x(1 - x)$ [duplicate]

This is the so-called logistic equation, which occurs often in population dynamics and many other contexts. There's a trick which works for this particular equation and is much simpler than separation of variables (in my opinion): change variables to $y(t)=1/p(t)$. Then the nonlinear equation for $p$ turns into an inhomogeneous linear equation for $y$, which can be solved immediately by the usual "homogeneous + particular solution" method (the homogeneous solution is an exponential, and the particular solution is a constant). Since this is tagged as homework, I'll let you have a go at the details yourself.

HINT: The method we can use here is called Separation of Variables. Take all the $p$'s to one side and $t$'s to the other, then integrate both sides like this: $$ \int \frac{dp}{10p(1-p)} = \int dt $$ and now integrate both sides. The right hand side is simply $ t+ C$ where C is some constant. To integrate the left hand side, use partial fractions. Set $$ \frac{1}{10p(1-p) } = \frac{A}{p} + \frac{B}{1-p} $$ and solve for $A $ and $B$, then both terms are easily integrated in terms of natural logs. After that, you will be able to solve for $p$ in terms of $t$ and some constant. Feel free to ask for more help if you need it!

From your comment, it looks you have been able to integrate correctly, following Ragib's Hint and Gourtaur comment. But now your problem is (to finish the solution) to express $p(t)$. This rest part is a simple algebra. Let me express $p(t)$ in terms of $t$:

$\frac{p}{1-p}=e^{10t+10c}=e^{10t}.e^{10c}=k.e^{10t}$ (where $k=e^{10c}$ is a new constant)

$\Rightarrow \frac{p}{(1-p)+p}=\frac{ke^{10t}}{1+ke^{10t}}$ (I applied $\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b+a}=\frac{c}{d+c}$. You can just multiply both sides by $(1-p)$, or cross-multiply and solve for $p$)

$\Rightarrow p=p(t)=\frac{1}{1+k'e^{-10t}}$ (dividing numerator and denominator of the fraction on RHS by $ke^{10t}$ and writing $k'=\frac{1}{k}$)

Now, from using the condition $p(0)=0.1=\frac{1}{10}$, we get $\frac{1}{10}=\frac{1}{1+k'}\Rightarrow k'=9$

Hence, you get $p(t)$ and when $t\to\infty$, $e^{-10t}$ tends to what?...