If ( e ^ 2 x +2 e ^ x - e ^ - x -1 ) e ^ ( e ^ x + e ^ - x ) d x … - Vidyadip

Question

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 integration, then $g (0)$ is equal to

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Answer

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

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

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

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

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

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

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

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