$$\frac{1}{dc} = \frac{dx}{εx}$$
That's what I was left with when I was finding the capacitance of a parallel plate capacitor with non-uniform dielectric in it. I just needed to find $c$ (capacitance) for this problem where the total distance ($D$) was also given.
But I don't think this integration which is $$∫\frac{1}{dc} = ∫\frac{dx}{εx}$$ [$ε$ is constant] can be solved. Can it be? Or am I doing something wrong?
Suppose we have a parallel plate capacitor with area $A$ and thickness $d$. We will establish a coordinate system in which the plates are on the planes $x=0$ and $x=d$. Assume that the permittivity varies as a function of $x$ so that $\epsilon = \epsilon(x)$.
Now, partition the interval $[0,d]$ into $N$ equal-sized subintervals of size $\Delta x=d/N$. The $n$th subinterval in $[x_{n-1},x_n]$ where $x_0=0$ and $x_N=d$.
For large $N$, the capacitance $C_n$ of the $n$th subinterval can be approximated by
$$C_n\approx \frac{\epsilon(x_n)A}{\Delta x}$$
We know that the aggregate structure is comprised of a series arrangement of the capacitors $C_n$. Therefore the total capacitance $C$ satisfies the equation
$$\begin{align} \frac{1}{C}&=\sum_{n=1}^N\frac{1}{C_n}\\\\ &\approx \sum_{n=1}^N\frac{1}{\epsilon(x_n)A}\Delta x \tag 1 \end{align}$$
If we take the limit as $N\to \infty$, and recognize that the right-hand side of $(1)$ is a Riemann sum, then we obtain
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{C}=\int_0^d \frac{1}{\epsilon(x)A}\,dx}$$