Question

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

• A. $2$
• B. $e^2$
• C. $e$
• D. $1$

Answer 1

Answer: (A)

$e^{2x}+2e^x-e^{-x}-1$
$=e^x\left(e^x+1\right)-e^{-x}\left(e^x+1\right)+e^x$
$=\left[\left(e^x+1\right)\left(e^x-e^{-x}\right)+e^x\right]$
so $I=\int \left(e^x+1\right)\left(e^x-e^{-x}\right)e^{e^x+e^{-x}}+\int e^x\cdot e^{e^x+e^{-x}}dx$
$=(e^x+1)e^{e^x+e^{-x}}-\int e^x.e^{e^x+e^{-x}}dx+\int e^x.e^{e^x+e^{-x}}dx$
$=\left(e^x+1\right)e^{e^x+e^{-1}}+C$
$\therefore g(x)=e^x+1$
$\Rightarrow g(0)=2$