Question
यदि $\int\left(e^{2 x}+2 e^{x}-e^{-x}-1\right) e^{\left(e^{x}+e^{-x}\right)} d x$ $= g ( x ) e ^{\left( e ^{ x }+ e ^{- x }\right)}+ c$ है, जहाँ $c$ एक समाकलन अचर है, तो $g (0)$ बराबर है
$= e ^{ x }\left( e ^{ x }+1\right)- e ^{- x }\left( e ^{ x }+1\right)+ e ^{ x }$
$=\left[\left( e ^{ x }+1\right)\left( e ^{ x }- e ^{- x }\right)+ e ^{ x }\right]$
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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