AB = x tan $\theta$, BC = d - x tan $\theta$,,

Capacitance of the strip is   

$\begin{array}{l}\mathrm{dC}=\frac{{\mathrm{\epsilon }}_0{\mathrm{A}}_{\mathrm{s}}}{\frac{\mathrm{AB}}{{\mathrm{K}}_1}+\frac{\mathrm{BC}}{{\mathrm{K}}_2}}\\ \left.\Rightarrow {\int }_0^{\mathrm{C}} \mathrm{dC}={\int }_0^{\mathrm{\ell }} \frac{{\mathrm{\epsilon }}_0\mathrm{bdx}}{\left(\frac{\mathrm{xtan}\mathrm{\theta }}{{\mathrm{K}}_1}+\frac{\mathrm{d}-\mathrm{xtan}\mathrm{\theta }}{{\mathrm{K}}_2}\right)}\right]\\ \Rightarrow \mathrm{C}=\frac{{\mathrm{\epsilon }}_0{\mathrm{bK}}_1{\mathrm{K}}_2}{\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)\tan \mathrm{\theta }}{\log }_{\mathrm{e}}\left[\frac{{\mathrm{K}}_1\mathrm{d}+\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)l\tan \mathrm{\theta }}{{\mathrm{K}}_1\mathrm{d}}\right]\end{array}$

But $\tan \mathrm{\theta }=\frac{\mathrm{d}}{l}$ and A = l x b

$\Rightarrow \mathrm{C}=\frac{{\mathrm{\epsilon }}_0{\mathrm{AK}}_1{\mathrm{K}}_2}{\left({\mathrm{K}}_2-{\mathrm{K}}_1\right)\mathrm{d}}{\log }_{\mathrm{e}}\left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{{\mathrm{\epsilon }}_0{\mathrm{K}}_1{\mathrm{K}}_2{\mathrm{a}}^2\ln \left(\frac{{\mathrm{K}}_1}{{\mathrm{K}}_2}\right)}{\left({\mathrm{K}}_1-{\mathrm{K}}_2\right)\mathrm{d}}$