If a continuous function `f` on `[0,a]` satisfies `f(x)f(a-x)=1,agt0`, then find the value of `int_(0)^(a)(dx)/(1+f(x))`.

Explanation

To solve the given problem, we need to evaluate the integral:

\[ I = \int_0^a \frac{dx}{1 + f(x)} \]

where \( f(x) \) is a continuous function satisfying the condition \( f(x) f(a - x) = 1 \) for \( x \in [0, a] \).

### Step 1: Use the property of the function
From the given property, we can express \( f(a - x) \) in terms of \( f(x) \):
\[ f(a - x) = \frac{1}{f(x)} \]

### Step 2: Rewrite the integral
Now, we can rewrite the integral \( I \) by substituting \( x \) with \( a - x \):
\[ I = \int_0^a \frac{dx}{1 + f(a - x)} = \int_0^a \frac{dx}{1 + \frac{1}{f(x)}} \]

### Step 3: Simplify the integral
Now simplify the expression:
\[ \frac{1}{1 + \frac{1}{f(x)}} = \frac{f(x)}{f(x) + 1} \]
Thus, we have:
\[ I = \int_0^a \frac{f(x)}{f(x) + 1} \, dx \]

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_0^a \frac{dx}{1 + f(x)} \)
2. \( I = \int_0^a \frac{f(x)}{f(x) + 1} \, dx \)

Adding these two equations:
\[ 2I = \int_0^a \left( \frac{1}{1 + f(x)} + \frac{f(x)}{f(x) + 1} \right) dx \]

### Step 5: Simplify the combined integral
The sum inside the integral simplifies as follows:
\[ \frac{1}{1 + f(x)} + \frac{f(x)}{f(x) + 1} = \frac{1 + f(x)}{1 + f(x)} = 1 \]
Thus, we have:
\[ 2I = \int_0^a 1 \, dx = a \]

### Step 6: Solve for \( I \)
Now, solving for \( I \):
\[ I = \frac{a}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^a \frac{dx}{1 + f(x)} = \frac{a}{2} \]