`int(e^(2x)+2e^x-e^(-x)-1)e^(e^x+e^(-x))dx=g(x)e^(e^x+e^(-x))` , then find `g(0)`.

Answer

To solve the given integral equation and find \( g(0) \), we start with the equation:

\[ \int (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})} \, dx = g(x)e^{(e^x + e^{-x})} \]

### Step 1: Differentiate both sides
We differentiate both sides with respect to \( x \). The left-hand side becomes:

\[ \frac{d}{dx} \left( \int (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})} \, dx \right) = (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})} \]

For the right-hand side, we apply the product rule:

\[ \frac{d}{dx} \left( g(x)e^{(e^x + e^{-x})} \right) = g'(x)e^{(e^x + e^{-x})} + g(x) \frac{d}{dx}(e^{(e^x + e^{-x})}) \]

Using the chain rule, we find:

\[ \frac{d}{dx}(e^{(e^x + e^{-x})}) = e^{(e^x + e^{-x})}(e^x - e^{-x}) \]

Thus, the right-hand side becomes:

\[ g'(x)e^{(e^x + e^{-x})} + g(x)e^{(e^x + e^{-x})}(e^x - e^{-x}) \]

### Step 2: Set the two sides equal
Now we equate the two sides:

\[ (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})} = g'(x)e^{(e^x + e^{-x})} + g(x)e^{(e^x + e^{-x})}(e^x - e^{-x}) \]

We can cancel \( e^{(e^x + e^{-x})} \) from both sides (assuming it is non-zero):

\[ e^{2x} + 2e^x - e^{-x} - 1 = g'(x) + g(x)(e^x - e^{-x}) \]

### Step 3: Rearranging the equation
Rearranging gives us:

\[ g'(x) + g(x)(e^x - e^{-x}) = e^{2x} + 2e^x - e^{-x} - 1 \]

### Step 4: Solve the differential equation
This is a first-order linear differential equation in \( g(x) \). We can solve it using an integrating factor or by inspection.

Let’s assume \( g(x) = e^x + C \) for some constant \( C \). Then:

\[ g'(x) = e^x \]

Substituting \( g(x) \) and \( g'(x) \) into the equation:

\[ e^x + (e^x + C)(e^x - e^{-x}) = e^{2x} + 2e^x - e^{-x} - 1 \]

### Step 5: Simplifying
Expanding the left side:

\[ e^x + (e^x + C)(e^x - e^{-x}) = e^x + e^{2x} - 1 + C(e^x - e^{-x}) \]

Setting the left side equal to the right side:

\[ e^x + e^{2x} - 1 + C(e^x - e^{-x}) = e^{2x} + 2e^x - e^{-x} - 1 \]

### Step 6: Comparing coefficients
From here, we can compare coefficients of \( e^x \) and \( e^{-x} \):

1. Coefficient of \( e^{2x} \): \( 1 = 1 \) (satisfied)
2. Coefficient of \( e^x \): \( 1 + C = 2 \) → \( C = 1 \)
3. Coefficient of \( e^{-x} \): \( -C = -1 \) → \( C = 1 \)

Thus, we find:

\[ g(x) = e^x + 1 \]

### Step 7: Find \( g(0) \)
Now we can find \( g(0) \):

\[ g(0) = e^0 + 1 = 1 + 1 = 2 \]

### Final Answer:
Thus, the value of \( g(0) \) is:

\[ \boxed{2} \]