1. Model the capacitor as many thin dielectric slabs in series

Since the permittivity varies along the separation direction $x$, we divide the capacitor into thin slices of thickness $dx$.

For a thin slice at position $x$: $dC = \frac{\varepsilon(x)A}{dx}$ But these slices are arranged in series along the field direction, so we must add their inverses: $d\left(\frac{1}{C}\right)=\frac{dx}{\varepsilon(x)A}$ Hence, $\frac{1}{C}=\frac{1}{A}\int_0^d \frac{dx}{\varepsilon(x)}$

2. Use the given permittivity function

Given, $\varepsilon(x)=\varepsilon_0+kx \quad \text{for } 0<x\le \frac d2$ and $\varepsilon(x)=\varepsilon_0+k(d-x) \quad \text{for } \frac d2\le x\le d$

Therefore, $\frac{1}{C}=\frac{1}{A}\left[\int_0^{d/2}\frac{dx}{\varepsilon_0+kx}+\int_{d/2}^{d}\frac{dx}{\varepsilon_0+k(d-x)}\right]$

3. Evaluate the first integral

$I_1=\int_0^{d/2}\frac{dx}{\varepsilon_0+kx}$ Let $u=\varepsilon_0+kx \Rightarrow du=k\,dx$ So, $I_1=\frac{1}{k}\ln(\varepsilon_0+kx)\Big|_0^{d/2}$ $I_1=\frac{1}{k}\ln\left(\frac{\varepsilon_0+kd/2}{\varepsilon_0}\right)$

4. Evaluate the second integral

$I_2=\int_{d/2}^{d}\frac{dx}{\varepsilon_0+k(d-x)}$ Let $u=\varepsilon_0+k(d-x) \Rightarrow du=-k\,dx$ Then, $I_2=\frac{1}{k}\ln\left(\frac{\varepsilon_0+kd/2}{\varepsilon_0}\right)$

So both integrals are equal.

Thus, $\frac{1}{C}=\frac{1}{A}\left(I_1+I_2\right)=\frac{2}{Ak}\ln\left(\frac{\varepsilon_0+kd/2}{\varepsilon_0}\right)$

Rewrite the logarithm: $\frac{\varepsilon_0+kd/2}{\varepsilon_0}=\frac{2\varepsilon_0+kd}{2\varepsilon_0}$ Hence, $\frac{1}{C}=\frac{2}{Ak}\ln\left(\frac{2\varepsilon_0+kd}{2\varepsilon_0}\right)$

Therefore, $C=\frac{kA}{2\ln\left(\frac{2\varepsilon_0+kd}{2\varepsilon_0}\right)}$

5. Match with the options

This matches Option B: $\boxed{\frac{kA}{2\ln\left(\frac{2\varepsilon_0+kd}{2\varepsilon_0}\right)}}$

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.