`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. Specifically, we can use the substitution \( x \to 2 - x \).

### Step 1: Substitute \( x \) with \( 2 - x \)

Using the substitution \( x = 2 - u \), we have:

\[ dx = -du. \]

When \( x = 0 \), \( u = 2 \), and when \( x = 2 \), \( u = 0 \). Thus, the integral becomes:

\[ I = \int_{2}^{0} \frac{e^{f(2-u)}}{e^{f(2-u)} + e^{f(u)}} (-du) = \int_{0}^{2} \frac{e^{f(2-u)}}{e^{f(2-u)} + e^{f(u)}} \, du. \]

### Step 2: Rewrite the integral

Now, we can rewrite the integral as:

\[ I = \int_{0}^{2} \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} \, dx. \]

### Step 3: Add the two expressions for \( I \)

Now we have two expressions for \( I \):

1. \( I = \int_{0}^{2} \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} \, dx \) (original)
2. \( I = \int_{0}^{2} \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} \, dx \) (after substitution)

Adding these two equations gives:

\[ 2I = \int_{0}^{2} \left( \frac{e^{f(x)}}{e^{f(x)} + e^{f(2-x)}} + \frac{e^{f(2-x)}}{e^{f(2-x)} + e^{f(x)}} \right) dx. \]

### Step 4: Simplify the expression

The sum inside the integral simplifies to:

\[ \frac{e^{f(x)} + e^{f(2-x)}}{e^{f(x)} + e^{f(2-x)}} = 1. \]

Thus, we have:

\[ 2I = \int_{0}^{2} 1 \, dx = 2. \]

### Step 5: Solve for \( I \)

Dividing both sides by 2 gives:

\[ I = 1. \]

### Final Answer

Thus, the value of the integral \( I \) is:

\[ \boxed{1}. \]