Prove the following


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 :

  1. 2

  2. $\mathrm{e}^{2}$

  3. $\mathrm{e}$

  4. 1

Correct Option: 1


$e^{2 x}+2 e^{x}-e^{-x}-1$



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

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

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

$\therefore \mathrm{g}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}+1 \Rightarrow \mathrm{g}(0)=2$

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