Question

If $\int (e^{2x}+2e^x-e^{-x}-1)e^{(e^x+e^{-x})}dx=g\left(x\right)e^{(e^x+e^{-x})}+c$, where $c$ is a constant of integration, then $g\left(0\right)$ is equal to :

• A. $2$
• B. $e$
• C. $1$
• D. $e^2$

Answer 1

The correct option is A $2$

$e^{2x}+2e^x-e^{-x}-1$
$=e^x(e^x+1)-e^{-x}(e^{-x}+1)+e^x$
$=\left[\right(e^x+1\left)\right(e^x-e^{-x})+e^x]\cdots \left(i\right)$
$I=\int (e^{2x}+2e^x-e^{-x}-1)e^{(e^x+e^{-x})}dx$
$I=\int \left[\right(e^x+1\left)\right(e^x-e^{-x})+e^x]e^{(e^x+e^{-x})}dx$
$I=\int (e^x+1)(e^x-e^{-x})e^{(e^x+e^{-x})}dx+\int e^x{e}^{(e^x+e^{-x})}dx$
Applying By-parts in first integral
$I=(e^x+1)e^{(e^x+e^{-x})}-\int e^x{e}^{(e^x+e^{-x})}dx+\int e^x{e}^{(e^x+e^{-x})}dx$
$I=(e^x+1)e^{(e^x+e^{-x})}+C$
$\therefore g\left(x\right)=e^x+1$
$\Rightarrow g\left(0\right)=1+1=2$