Key Concepts and Formulas

• Indefinite Integration: The process of finding a function whose derivative is a given function.
• Substitution Method: A technique used to simplify integrals by substituting a part of the integrand with a new variable.
• Properties of Exponentials: $e^{a+b} = e^a \cdot e^b$ and the derivative of $e^x$ is $e^x$.

Step-by-Step Solution

Step 1: Rewrite the integral and separate terms. We begin by rewriting the given integral to group terms strategically for simplification and apply the property of integrals of sums. $I = \int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} = \int {\left( {{e^{2x}} + {e^x} + {e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}$ $I = \int {\left( {{e^{2x}} + {e^x} - 1 + {e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}$ $I = \int {\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} + \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}$

Step 2: Manipulate the first integral to prepare for substitution. We divide and multiply the first integral by $e^x$ to modify it into a form suitable for substitution. This algebraic manipulation doesn't change the value of the integral. $I = \int {\frac{{\left( {{e^{2x}} + {e^x} - 1} \right)}}{{{e^x}}}{e^x}{e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} + \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}$ $I = \int {\left( {{e^x} + 1 - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx} + \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}$

Step 3: Apply substitution to the first integral. Let $t = {e^x} + {e^{ - x}} + x$. Then, $\frac{dt}{dx} = {e^x} - {e^{ - x}} + 1$, so $dt = \left( {{e^x} - {e^{ - x}} + 1} \right)dx$. Substituting these into the first integral, we get: $I_1 = \int {\left( {{e^x} + 1 - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx} = \int {{e^t}dt} = {e^t} + c_1 = {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}} + c_1$

Step 4: Apply substitution to the second integral. Let $u = {e^x} + {e^{ - x}}$. Then, $\frac{du}{dx} = {e^x} - {e^{ - x}}$, so $du = \left( {{e^x} - {e^{ - x}}} \right)dx$. Substituting these into the second integral, we get: $I_2 = \int {\left( {{e^x} - {e^{ - x}}} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} = \int {{e^u}du} = {e^u} + c_2 = {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c_2$

Step 5: Combine the results of the two integrals. $I = I_1 + I_2 = {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}} + {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + C$ $I = {e^{\left( {{e^x} + {e^{ - x}}} \right)}}{e^x} + {e^{\left( {{e^x} + {e^{ - x}}} \right)}} + C$ $I = {e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right) + C$

Step 6: Identify g(x) and calculate g(0). We are given that $I = g(x){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$. Comparing this with our result, we can identify $g(x) = {e^x} + 1$. Now we need to find $g(0)$. $g(0) = {e^0} + 1 = 1 + 1 = 2$

Common Mistakes & Tips

• Careless Algebra: Be very careful with algebraic manipulations, especially when dealing with exponential functions. A small error can propagate through the entire solution.
• Choosing the Right Substitution: The key is to identify a part of the integrand whose derivative (or a multiple thereof) is also present.
• Don't Forget the Constant of Integration: Always remember to add the constant of integration, 'C', when evaluating indefinite integrals.

Summary

We rewrote the integral and separated it into two parts. We then used the substitution method to solve each part individually. After obtaining the solutions for each integral, we combined them to find the overall solution. Finally, we identified the function $g(x)$ and calculated $g(0)$, which equals 2.

Final Answer

The final answer is \boxed{2}, which corresponds to option (B).