If int (e^{2x} + 2e^{x} - e^{-x} - 1) e^{(e^{ | vedclass

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(A) Let $I = \int (e^{2x} + 2e^{x} - e^{-x} - 1) e^{(e^{x} + e^{-x})} dx$.We can rewrite the integrand as:$(e^{2x} + 2e^{x} - e^{-x} - 1) = e^{x}(e^{x

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$2$

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(A) Let $I = \int (e^{2x} + 2e^{x} - e^{-x} - 1) e^{(e^{x} + e^{-x})} dx$.We can rewrite the integrand as:$(e^{2x} + 2e^{x} - e^{-x} - 1) = e^{x}(e^{x} + 1) - e^{-x}(e^{x} + 1) + e^{x} = (e^{x} + 1)(e^{x} - e^{-x}) + e^{x}$.So,$I = \int (e^{x} + 1)(e^{x} - e^{-x}) e^{(e^{x} + e^{-x})} dx + \int e^{x} e^{(e^{x} + e^{-x})} dx$.For the first integral,let $u = e^{x} + e^{-x}$,then $du = (e^{x} - e^{-x}) dx$. However,this is not quite right. Let's use integration by parts on the first part.Let $f(x) = e^{x} + 1$ and $h'(x) = (e^{x} - e^{-x}) e^{(e^{x} + e^{-x})}$.Then $h(x) = e^{(e^{x} + e^{-x})}$.Using integration by parts $\int f h' = fh - \int f' h$:$I = (e^{x} + 1) e^{(e^{x} + e^{-x})} - \int e^{x} e^{(e^{x} + e^{-x})} dx + \int e^{x} e^{(e^{x} + e^{-x})} dx$.$I = (e^{x} + 1) e^{(e^{x} + e^{-x})} + c$.Comparing this with $g(x) e^{(e^{x} + e^{-x})} + c$,we get $g(x) = e^{x} + 1$.Therefore,$g(0) = e^{0} + 1 = 1 + 1 = 2$.

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 equal to

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