`I=int_(0)^(2)(e^(f(x)))/(e^(f(x))+e^(f(2-x)))dx` is equal to

Answer

To solve the integral \( I = \int_{0}^{2} \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} \, dx \), we can use a property of definite integrals. Here are the detailed steps:

### Step 1: Rewrite the Integral
We start with the given integral:
\[ I = \int_{0}^{2} \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} \, dx \]

### Step 2: Change of Variable
We can use the substitution \( u = 2 - x \). Then, \( du = -dx \). When \( x = 0 \), \( u = 2 \), and when \( x = 2 \), \( u = 0 \). Therefore, we can rewrite the integral as:
\[ I = \int_{2}^{0} \frac{e^{f(2-u)}}{e^{f(2-u)} + e^{f(u)}} (-du) = \int_{0}^{2} \frac{e^{f(2-u)}}{e^{f(2-u)} + e^{f(u)}} \, du \]

### Step 3: Simplify the New Integral
Now we can denote this new integral as \( J \):
\[ J = \int_{0}^{2} \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} \, dx \]

### Step 4: Add the Two Integrals
Now we have two expressions for \( I \):
\[ I = \int_{0}^{2} \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} \, dx \]
\[ J = \int_{0}^{2} \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} \, dx \]
Adding these two integrals, we get:
\[ I + J = \int_{0}^{2} \left( \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} + \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} \right) dx \]

### Step 5: Simplify the Sum
Notice that:
\[ \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} + \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} = 1 \]
Thus, we have:
\[ I + J = \int_{0}^{2} 1 \, dx = 2 \]

### Step 6: Relate \( I \) and \( J \)
Since \( I = J \), we can write:
\[ 2I = 2 \implies I = 1 \]

### Conclusion
Thus, the value of the integral is:
\[ \boxed{1} \]