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?...