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

  1. (1) $e$

  2. (2) $e^{2}$

  3. (3) 1

  4. (4) 2

Correct Option: , 4


$\int\left(e^{2 x}+2 e^{x}-e^{-x}-1\right) \cdot e^{\left(e^{x}+e^{-x}\right)} d x$

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

$=\int e^{x}\left(e^{x}+1-e^{-x}\right) \cdot e^{\left(e^{x}+e^{-x}\right)} d x+e^{\left(e^{x}+e^{-x}\right)}$

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

Let $e^{x}+e^{-x}+x=t \Rightarrow\left(e^{x}+e^{-x}+1\right) d x=d t$

$=\int e^{t} d t+e^{\left(e^{x}+e^{-x}\right)}=e^{t}+e^{\left(e^{x}+e^{-x}\right)}+C$


$=\left(e^{x}+1\right) \cdot e^{\left(e^{x}+e^{-x}\right)}+C$

So, $g(x)=1+e^{x}$ and $g(0)=2$

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