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