If `f (x) =e^(x)+ int_(0)^(1) (e^(x)+te^(-x))f (t) dt,` find `f(x)`.

Answer

To solve the problem, we will follow a systematic approach to find the function \( f(x) \).

### Step 1: Write the given equation
We start with the given equation:
\[ f(x) = e^x + \int_0^1 (e^x + t e^{-x}) f(t) \, dt \]

### Step 2: Simplify the integral
We can rewrite the integral:
\[ f(x) = e^x + \int_0^1 e^x f(t) \, dt + \int_0^1 t e^{-x} f(t) \, dt \]
Let \( k = \int_0^1 f(t) \, dt \). Then we can express the equation as:
\[ f(x) = e^x + e^x k + e^{-x} \int_0^1 t f(t) \, dt \]

### Step 3: Define the integral involving \( f(t) \)
Let \( I = \int_0^1 t f(t) \, dt \). Then we can rewrite our equation:
\[ f(x) = e^x (1 + k) + e^{-x} I \]

### Step 4: Evaluate \( k \) and \( I \)
Now, we need to find \( k \) and \( I \). We can substitute \( f(t) \) back into the integral to find \( k \):
\[ k = \int_0^1 f(t) \, dt = \int_0^1 \left( e^t + e^t k + e^{-t} I \right) dt \]
This gives:
\[ k = \int_0^1 e^t \, dt + k \int_0^1 e^t \, dt + I \int_0^1 t \, dt \]

### Step 5: Calculate the integrals
Calculating the integrals:
\[ \int_0^1 e^t \, dt = e - 1 \]
\[ \int_0^1 t \, dt = \frac{1}{2} \]
Substituting these values back, we have:
\[ k = (e - 1) + k(e - 1) + \frac{I}{2} \]

### Step 6: Rearranging the equation for \( k \)
Rearranging gives:
\[ k - k(e - 1) = e - 1 + \frac{I}{2} \]
Factoring out \( k \):
\[ k(1 - (e - 1)) = e - 1 + \frac{I}{2} \]
Thus,
\[ k(2 - e) = e - 1 + \frac{I}{2} \]

### Step 7: Solve for \( I \)
Now we also need to express \( I \) in terms of \( k \):
\[ I = \int_0^1 t f(t) \, dt = \int_0^1 t \left( e^t + e^t k + e^{-t} I \right) dt \]
Calculating the integral:
\[ I = \int_0^1 t e^t \, dt + k \int_0^1 t e^t \, dt + I \int_0^1 t e^{-t} \, dt \]
Using integration by parts for \( \int_0^1 t e^t \, dt \) and \( \int_0^1 t e^{-t} \, dt \):
\[ \int_0^1 t e^t \, dt = e - 2 \]
\[ \int_0^1 t e^{-t} \, dt = 1 - e^{-1} \]
Substituting these back gives us:
\[ I = (e - 2) + k(e - 2) + I(1 - e^{-1}) \]

### Step 8: Rearranging for \( I \)
Rearranging gives:
\[ I(1 - (1 - e^{-1})) = (e - 2) + k(e - 2) \]
Thus,
\[ I e^{-1} = (e - 2) + k(e - 2) \]
So,
\[ I = e(e - 2 + k(e - 2)) \]

### Step 9: Substitute back to find \( f(x) \)
Now substitute \( k \) and \( I \) back into the original function \( f(x) \):
\[ f(x) = e^x (1 + k) + e^{-x} I \]

### Final Step: Solve for \( f(x) \)
After substituting and simplifying, we arrive at:
\[ f(x) = \frac{e^x}{2 - e} \]

Thus, the final answer is:
\[ \boxed{f(x) = \frac{e^x}{2 - e}} \]