Your questions aren't entirely clear to me, so please accept my apologies if I wander just a bit. I'll avoid using integrating factors below, as well, since it's not necessary for this problem.
For ideal capacitors (few are), the basic idea is \$Q=C\:V\$. Taking the full derivative, this is \$\text{d} Q = C\:\text{d} V+V\:\text{d}C\$. But it's usually assumed that \$\text{d}C=0\$ -- that there is no change in capacitance, since it is a fixed value in an ideal capacitor. So \$\text{d} Q = C\:\text{d} V\$. You can now introduce time by simply dividing both sides by \$\text{d} t\$, giving \$\frac{\text{d} Q}{\text{d} t} = C\:\frac{\text{d} V}{\text{d} t}=I\$. (The ad-hoc introduction of the infinitesimal of time can be done where it pleases.)
Your KVL loop equation is:
$$\begin{align*}
V &= R\:I_{\left(t\right)}+\frac{1}{C}\int I_{\left(t\right)}\:\text{d}t\tag{1}\\\\
\text{d} V&=R\:\text{d}I_{\left(t\right)}+\frac{I_{\left(t\right)}}{C}\text{d} t=0\tag{2}\\\\
\text{d}I_{\left(t\right)}&=-\frac{I_{\left(t\right)}}{R\,C}\text{d} t\tag{3}\\\\
\frac{\text{d}I_{\left(t\right)}}{I_{\left(t\right)}}&=\frac{-1}{R\,C}\text{d} t\tag{4}\\\\
\int\frac{\text{d}I_{\left(t\right)}}{I_{\left(t\right)}}&=\frac{-1}{R\,C}\int\text{d} t\tag{5}\\\\
\operatorname{ln}\left(I_{\left(t\right)}\right)&=\frac{-t}{R\,C} + A_0\tag{6}\\\\
I_{\left(t\right)}&=e^{\left[\frac{-t}{R\,C} + A_0\right]}=e^{A_0}e^{\frac{-t}{R\,C}}=A\,e^{\frac{-t}{R\,C}}\tag{7}
\end{align*}$$
Where \$A\$ is the constant of integration you need to worry about, now. At \$t=0\$, it must be that \$A=I_{t=0}\$. If the voltage across the capacitor at \$t=0\$ is \$V_C=0\:\text{V}\$, then it follows that \$A=\frac{V}{R}\$. So the result must be:
$$I_{\left(t\right)}=\frac{V}R\,e^{\frac{-t}{R\,C}}\tag{Series Current}\label{E1}$$
And,
$$\begin{align*}
V_{C\left(t\right)} &= \frac{1}{C}\int I_{\left(t\right)}\:\text{d}t\tag{8}\\\\
&=\frac{1}{C}\int \frac{V}R\,e^{\frac{-t}{R\,C}}\:\text{d}t\tag{9}\\\\
&=\frac{V}{R\,C}\int e^{\frac{-t}{R\,C}}\:\text{d}t\tag{10}\\\\
&=\frac{V}{R\,C}\left[-R\,C\,e^{\frac{-t}{R\,C}} + A_0\right]\tag{11}\\\\
&=V\left[\frac{A_0}{R\,C}-e^{\frac{-t}{R\,C}}\right]\tag{12}
\end{align*}$$
Again, at \$t=0\$ and assuming that \$V_{C\left(t=0\right)}=0\:\text{V}\$, it follows that \$A_0=R\,C\$. So:
$$V_{C\left(t\right)}=V\left[1-e^{\frac{-t}{R\,C}}\right]\label{E2}\tag{Capacitor Voltage}$$
That's it. The \$\ref{E1}\$ equation provides the current in both \$R\$ and \$C\$, since both must be the same. The \$\ref{E2}\$ equation provides the voltage that accumulates onto the capacitor over time.