A dielectric slab is introduced inside an uncharged capacitor of plate area A and plate separation L. The surface area of the slab is A and its thickness is slightly less than L. Let the surfaces of the slab that are facing the plates of the capacitor be called its two “faces”. The dielectric constant of the slab varies with distance x from one of its faces as: `K=K_(0)(1+(X)/(L))`, where `K_(0)` is a constant. After the insertion of the slab, the capacitance of the capacitor is `n((K_(0)epsilon_(0)A)/(Llog_(e)2))`. The value of n is _________.

Explanation

To solve the problem, we need to find the value of \( n \) in the expression for the capacitance of a capacitor with a dielectric slab inserted, where the dielectric constant varies with distance \( x \) from one of its faces.

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a parallel plate capacitor with plate area \( A \) and plate separation \( L \). A dielectric slab with thickness slightly less than \( L \) is inserted between the plates. The dielectric constant \( K \) varies with distance \( x \) from one of its faces as:
\[ K = K_0 \left(1 + \frac{x}{L}\right) \]

2. **Dividing the Slab into Differential Elements**:
We consider a differential element of thickness \( dx \) at a distance \( x \) from one of the capacitor plates. The capacitance \( dC \) of this differential element can be expressed as:
\[ dC = \frac{\epsilon_0 K A}{dx} \]
Here, \( K \) is the dielectric constant at distance \( x \), which we substitute:
\[ dC = \frac{\epsilon_0 A}{dx} K_0 \left(1 + \frac{x}{L}\right) \]

3. **Expressing \( dC \) in Terms of \( dx \)**:
Since the thickness of the slab is \( dx \), we rewrite the expression for \( dC \):
\[ dC = \frac{\epsilon_0 A K_0}{dx} \left(1 + \frac{x}{L}\right) \]

4. **Finding the Total Capacitance**:
The total capacitance \( C \) can be found by integrating \( dC \) over the thickness of the slab from \( 0 \) to \( L \):
\[ \frac{1}{C} = \int_0^L \frac{1}{dC} \]
Since \( dC \) is expressed in terms of \( dx \), we need to integrate:
\[ \frac{1}{C} = \int_0^L \frac{dx}{\epsilon_0 A K_0 \left(1 + \frac{x}{L}\right)} \]

5. **Simplifying the Integral**:
The integral can be simplified:
\[ \frac{1}{C} = \frac{1}{\epsilon_0 A K_0} \int_0^L \frac{dx}{1 + \frac{x}{L}} \]
Substituting \( u = 1 + \frac{x}{L} \) gives \( dx = L du \), changing the limits accordingly:
\[ \frac{1}{C} = \frac{L}{\epsilon_0 A K_0} \int_1^{1 + \frac{L}{L}} \frac{du}{u} = \frac{L}{\epsilon_0 A K_0} \left[ \ln u \right]_1^2 \]

6. **Evaluating the Integral**:
Evaluating the integral:
\[ \frac{1}{C} = \frac{L}{\epsilon_0 A K_0} \left( \ln 2 - \ln 1 \right) = \frac{L \ln 2}{\epsilon_0 A K_0} \]

7. **Finding the Capacitance**:
Thus, we find:
\[ C = \frac{\epsilon_0 A K_0}{L \ln 2} \]

8. **Comparing with Given Expression**:
The problem states that the capacitance after the insertion of the slab is:
\[ C = n \left( \frac{K_0 \epsilon_0 A}{L \ln 2} \right) \]
By comparing both expressions, we see that:
\[ n = 1 \]

### Final Answer:
The value of \( n \) is \( \boxed{1} \).