If $$\int (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})}\,dx = g(x)e^{(e^x + e^{-x})} + c$$, where $$c$$ is a constant of integration, then $$g(0)$$ is:

We begin by writing the given integral in a compact form. Put

$$$F(x)=e^{x}+e^{-x},\qquad\text{so that}\qquad e^{F(x)}=e^{\,(e^{x}+e^{-x})}.$$$

The statement of the problem is

$$$\int\bigl(e^{2x}+2e^{x}-e^{-x}-1\bigr)\,e^{F(x)}\,dx = g(x)\,e^{F(x)}+c,$$$

where $$c$$ is the (single) constant of integration. Define

$$I(x)=g(x)\,e^{F(x)}.$$

Because the left-hand side is an antiderivative of the integrand, differentiating both sides gives back the integrand. Using the product rule,

$$$\frac{d}{dx}\bigl[g(x)\,e^{F(x)}\bigr] = g'(x)\,e^{F(x)}+g(x)\,e^{F(x)}\,F'(x) = \bigl[g'(x)+g(x)\,F'(x)\bigr]e^{F(x)}.$$$

But the derivative must equal the integrand itself, so we must have

$$$\bigl[g'(x)+g(x)\,F'(x)\bigr]e^{F(x)} =\bigl(e^{2x}+2e^{x}-e^{-x}-1\bigr)e^{F(x)}.$$$

Cancelling the common positive factor $$e^{F(x)}$$ produces the first-order linear differential equation

$$$g'(x)+g(x)\,F'(x)=e^{2x}+2e^{x}-e^{-x}-1. \quad -(1)$$$

We now evaluate $$F'(x)$$ explicitly:

$$F'(x)=\frac{d}{dx}(e^{x}+e^{-x})=e^{x}-e^{-x}.$$

Substituting this into (1) gives

$$$g'(x)+\bigl(e^{x}-e^{-x}\bigr)\,g(x)=e^{2x}+2e^{x}-e^{-x}-1. \quad -(2)$$$

The right‐hand side contains the terms $$e^{2x},e^{x},e^{-x}$$ and a constant. A very economical guess is that $$g(x)$$ itself should involve only the two simplest exponentials $$e^{x}$$ and $$e^{0}=1$$. So let us try

$$g(x)=Ae^{x}+B,$$

where $$A$$ and $$B$$ are constants to be determined. We compute each piece in (2) separately.

First, the derivative of $$g(x)$$:

$$g'(x)=A\,e^{x}.$$

Second, the product $$\bigl(e^{x}-e^{-x}\bigr)\,g(x)$$:

$$$\bigl(e^{x}-e^{-x}\bigr)\,g(x) =\bigl(e^{x}-e^{-x}\bigr)\bigl(Ae^{x}+B\bigr) =A\,e^{2x}+B\,e^{x}-A\,e^{x}-B\,e^{-x}.$$$

Adding these two expressions, we obtain the left side of (2):

$$$\begin{aligned} g'(x)+\bigl(e^{x}-e^{-x}\bigr)\,g(x) &=A\,e^{x}+A\,e^{2x}+B\,e^{x}-A\,e^{x}-B\,e^{-x}\\[4pt] &=A\,e^{2x}+B\,e^{x}-B\,e^{-x}. \end{aligned}$$$

For this to match the right side of (2), namely $$e^{2x}+2e^{x}-e^{-x}-1,$$ we equate coefficients term by term:

• Coefficient of $$e^{2x}:\quad A=1.$$

• Coefficient of $$e^{x}:\quad B=2.$$

• Coefficient of $$e^{-x}:\quad -B=-1\;\;\Longrightarrow\;\;B=1.$$

The last two requirements conflict unless $$B=2$$ and $$B=1$$ simultaneously, which is impossible. Therefore $$g(x)=Ae^{x}+B$$ is not general enough. We refine our guess by adding one more term and set

$$g(x)=Ae^{x}+B+Ce^{-x}.$$

Now proceed again. First,

$$g'(x)=A\,e^{x}-C\,e^{-x}.$$

Next,

$$$\bigl(e^{x}-e^{-x}\bigr)\,g(x)= (e^{x}-e^{-x})(Ae^{x}+B+Ce^{-x}) =A\,e^{2x}+B\,e^{x}+C -A\,e^{x}-B\,e^{-x}-C\,e^{-2x}.$$$

Adding, we find

$$$\begin{aligned} g'(x)+\bigl(e^{x}-e^{-x}\bigr)\,g(x) &=(A\,e^{x}-C\,e^{-x})\\ &\quad+\bigl[A\,e^{2x}+B\,e^{x}+C -A\,e^{x}-B\,e^{-x}-C\,e^{-2x}\bigr]\\[4pt] &=A\,e^{2x}+(B\,e^{x})+C -C\,e^{-x}-C\,e^{-2x}. \end{aligned}$$$

We must match this with $$e^{2x}+2e^{x}-e^{-x}-1$$. Noticing that the target expression contains no $$e^{-2x}$$ term, we set

$$C=0.$$

That immediately wipes out the unwanted $$e^{-2x}$$ term and simplifies the comparison to

$$A\,e^{2x}+B\,e^{x}=e^{2x}+2e^{x}-e^{-x}-1.$$

Now the remaining $$e^{-x}$$ and constant terms still do not appear on the left, so the only way they can be produced is if we have in fact already chosen the correct structure and only need to re-check our algebra. A quicker route is to test the simplest imaginable specific choice, namely

$$g(x)=e^{x}+1.$$

Compute once more with this concrete $$g$$:

$$g'(x)=e^{x},$$

$$$g(x)\,F'(x)=(e^{x}+1)(e^{x}-e^{-x})=e^{2x}-e^{-x}+e^{x}-1.$$$

Adding these two gives

$$$g'(x)+g(x)\,F'(x) =e^{x}+\bigl(e^{2x}-e^{-x}+e^{x}-1\bigr) =e^{2x}+2e^{x}-e^{-x}-1,$$$

which is exactly the right side of (2). Hence the correct choice is

$$g(x)=e^{x}+1.$$

Finally, we evaluate at $$x=0$$:

$$g(0)=e^{0}+1=1+1=2.$$

Hence, the correct answer is Option D.