A parallel plate capacitor has plates of area A separated by distance d between them. It is filled with a dielectric which has a dielectric constant that varies as $$K(x) = K_0(1 + \alpha x)$$ where $$x$$ is the distance measured from one of the plates. If $$(\alpha d) \ll 1$$, the total capacitance of the system is best given by the expression:

Capacitance of an elemental differential slab of thickness $$dx$$ at distance $$x$$: 

$$dC = \frac{K(x) \varepsilon_0 A}{dx} = \frac{K_0(1 + \alpha x)\varepsilon_0 A}{dx}$$

Since elements at different $$x$$ are stacked in series across the plates:

$$\frac{1}{C} = \int_{0}^{d} \frac{1}{dC} = \int_{0}^{d} \frac{dx}{K_0 \varepsilon_0 A (1 + \alpha x)}$$

$$\frac{1}{C} = \frac{1}{K_0 \varepsilon_0 A \alpha} \left[ \ln(1 + \alpha x) \right]_0^d = \frac{\ln(1 + \alpha d)}{K_0 \varepsilon_0 A \alpha}$$

Using the Taylor expansion $$\ln(1 + z) \approx z - \frac{z^2}{2}$$ since $$\alpha d \ll 1$$:

$$\frac{1}{C} \approx \frac{1}{K_0 \varepsilon_0 A \alpha} \left( \alpha d - \frac{\alpha^2 d^2}{2} \right) = \frac{d}{K_0 \varepsilon_0 A} \left( 1 - \frac{\alpha d}{2} \right)$$

$$C = \frac{K_0 \varepsilon_0 A}{d} \left( 1 - \frac{\alpha d}{2} \right)^{-1}$$

Using the binomial approximation $$(1 - z)^{-1} \approx 1 + z$$: $$C = \frac{A K_0 \varepsilon_0}{d} \left( 1 + \frac{\alpha d}{2} \right)$$