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Q.
If $\int\left( e ^{2 x }+2 e ^{ x }- e ^{- x }-1\right) e ^{\left( c ^{ x }+ e ^{- x }\right)} dx$
$=g(x) e^{\left(e^{2}+e^{-x}\right)}+c,$ where $c$ is a constant of
integration, then $g (0)$ is equal to :
$e ^{2 x }+2 e ^{ x }- e ^{- x }-1$
$= e ^{ x }\left( e ^{ x }+1\right)- e ^{- x }\left( e ^{ x }+1\right)+ e ^{ x }$
$=\left[\left(e^{x}+1\right)\left(e^{x}-e^{-x}\right)+e^{x}\right]$
so $I=\int\left(e^{x}+1\right)\left(e^{x}-e^{-x}\right) e^{e^{x}+e^{-x}}+\int e^{x} \cdot e^{e^{x}+e^{-x}} d x$
$= (e^x + 1) e^{e^x + e^{-x}} - \int e^x .e^{e^x + e^{-x}} dx + \int e^x . e^{e^x + e^{-x}}dx$
$=\left( e ^{ x }+1\right) e ^{ e ^{x}+ e ^{-1}}+ C$
$\therefore g ( x )= e ^{x}+1$
$\Rightarrow g (0)=2$