Voltage divider equation except for capacitors?
I seriously apologize for the horrible formatting but I don't know LaTex OR MathJax and I have to leave town in the morning.  

You could solve this using voltage and current divider equations, but I find it easier to just consider the current flows. Let \$I_1, I_2, I_3\$ and \$I_4\$ denote the current through \$C_1, C_2,\$ etc., and let \$V_1, V_2,\$ etc., denote the voltage across the corresponding capacitors.  Also recall that \$V = IZ\$, where \$Z = 1/sC\$ for a capacitor in the Laplace domain.

Notice that \$I_s = I_1 = I_2+I_3 = I_2+I_4\$, since \$I_3 = I_4\$.

Now \$V_2 = V_3 + V_4\$ , so 
$$
\frac{I_2}{sC_2} = \frac{I_3}{sC_3} + \frac{I_4}{sC_4}
$$

but \$I_3 = I_4\$, so
$$
\begin{split}
\frac{I_2}{sC_2} &amp;= \frac{I_4}{sC_3} + \frac{I_4}{sC_4} \\
&amp;= \frac{I_4}{s}\left(\frac{1}{C_3} + \frac{1}{C_4}\right)\\
&amp;= \frac{I_4}{s}\frac{C_4+C_3}{C_3C_4}= I_4\frac{C_3+C_4}{sC_3C_4}
\end{split}
$$
and 
$$
I_2 = I_4\frac{C_2(C_3+C_4)}{(C_3C_4)}
$$
Now 
$$
\begin{split}
I_s &amp; = I_4 + I_2\\
   &amp; = I_4 + I_4\frac{C_2(C_3+C_4)}{C_3C_4}\\
   &amp; = I_4\left\{1 + \frac{C_2(C_3+C_4)}{C_3C_4}\right\}\\
   &amp; = I_4\frac{C_3C_4 + C_2(C_3+C_4)}{C_3C_4}
\end{split}
$$
so
$$
I_4 = I_s\frac{C_3C_4}{C_3C_4 + C_2(C_3+C_4)}
$$
Now 
$$
V_\mathrm{out} = V_4 = \frac{I_4}{sC_4}, 
$$
so
$$
\begin{split}
V_\mathrm{out} &amp;= I_s\frac{C_3C_4[C_3C_4 + C_2(C_3+C_4)]}{sC_4}\\
&amp; = I_s\frac{C_3}{s(C_3C_4 + C_2C_3 + C_2C_4)} = \frac{I_s}{sC}
\end{split}
$$
where
$$
C = \frac{C_2C_3 + C_2C_4 + C_3C_4}{C_3}
$$
Note that in the time domain, \$V_\mathrm{out}(t)\$ is \$1/C\$ times the integral of \$I_s(t)\$, so if \$I_s(t)\$ has any DC component, \$V_\mathrm{out}\$ will increase indefinitely until dielectric breakdown in one of the capacitors.

Hope I didn't make a mistake in all that, but you can see the general idea. :)