Let the competitive reactions be $A \stackrel{k_1}{\rightarrow}B$ and $A \stackrel{k_2}{\rightarrow}C$ of first order, we will determine $[A]$, $[B]$ and $[ C]$ as a function of time, assuming $[B]_0=[C]_0=0$. The variation of $[A]$ in time is given by the expression:
\begin{equation}\frac{d[A]}{dt}=-k_1[A]-k_2[A]\;\;\rightarrow \frac{d[A]}{dt}=(-k_1-k_2)[A]\end{equation}
Separating variables and integrating:
\begin{equation}[A]=[A]_0 e^{-( k_1+k_2)t} \label{ec1}\end{equation}
We now proceed to obtain the variation of $[B]$ with t
\begin{equation} \frac{d[B]}{dt}=k_1[A] \end{equation}
Substituting $[A]$ for $(\ref{ec1})$
\begin{equation} \frac{d[B]}{dt}=k_1[A]_0 e^{-(k_1+k_2)t} \end{equation}
Separating variables and integrating
\begin{equation} [B]=\left[\frac{k_1[A]_0}{-(k_1+k_2)}^{-(k_1+k_2)t}\right ]_{0}^{t} \end{equation}
Substituting the limits of integration
\begin{equation} [B]=\frac{k_1 [A]_0}{k_1 +k_2}\left[1-e^{ -(k_1+k_2)t}\right] \end{equation}
The variation of $[C]$ with t follows analogously from the differential equation $\frac{d[C]}{dt}= k_2[A]$
\begin{equation} [C]=\frac{k_2[A]_0}{k_1 +k_2}\left[1-e^{-(k_1+k_2)t}\right] \end{equation}
Dividing both concentrations at any inst before time we get:
\begin{equation} \frac{[C]}{[B]}=\frac{k_2}{k_1} \end{equation}