If ∫ e 2 x + 2 e x - e - x - 1 e e x + e - x d x = g ( x ) e e x + e - x + c , where c is a constant…

$I=\int \left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}dx$

$\Rightarrow I=\int \left(e^{2x}+e^x-1\right)e^{\left(e^x+e^{-x}\right)}dx+\int \left(e^x-e^{-x}\right)e^{e^x+e^{-x}}dx$

$\Rightarrow I=\int \left(e^x+1-e^{-x}\right)e^{e^x+e^{-x}+x}dx+e^{e^x+e^{-x}}+c$

$e^x+e^{-x}+x=u$

$\left(e^x-e^{-x}+1\right)dx=du$

$I=e^{e^x+e^{-x}+x}+e^{e^x+e^{-x}}+c=e^{e^x+e^{-x}}\left(e^x+1\right)+c$

then $g\left(x\right)=e^x+1$

$g\left(0\right)=2$