For , , , N, if (((x)/(e))^2x+((e)/(x))^2x) _exdx=(1)/( )((x)/(e))^ x-(1)/( )((e)/(x))^ x+…

Question

For $\alpha ,\beta ,\gamma ,\delta \in N$, if $\int ({(\frac{x}{e})}^{2x}+{(\frac{e}{x})}^{2x}){\log }_{e}xdx=\frac{1}{\alpha }{(\frac{x}{e})}^{\beta x}-\frac{1}{\gamma }{(\frac{e}{x})}^{\delta x}+C$

, where $e=\sum _{n=0}^{\infty}\frac{1}{n!}$ and $C$ is constant of integration, then $\alpha +2\beta +3\gamma -4\delta $ is equal to :

Accepted Answer

We have,

$\int ({(\frac{x}{e})}^{2x}+{(\frac{e}{x})}^{2x}){\log }_{e}xdx=\frac{1}{\alpha }{(\frac{x}{e})}^{\beta x}-\frac{1}{\gamma }{(\frac{e}{x})}^{\delta x}+C$

$\begin{gathered} {(\frac{x}{e})}^{2x} ={(\frac{{e}^{{\log }_{e}x}}{e})}^{2x}[\because x={e}^{{\log }_{e}x}] \ \ ={e}^{2(x{\log }_{e}x-x)}.........(i) $

${(\frac{e}{x})}^{2x}={(\frac{x}{e})}^{-2x}={e}^{-2(x{\log }_{e}x-x)}$ .........(ii)

Let,

$ I =\int [{(\frac{x}{e})}^{2x}+{(\frac{e}{x})}^{2x}]{\log }_{e}xdx \ \ =\int [{e}^{2(x{\log }_{e}x-x)}+{e}^{-2(x{\log }_{e}x-x)}]{\log }_{e}xdx $

Let $x{\log }_{e}x-x=t$

${\log }_{e}xdx=dt$

$ I =\int ({e}^{2t}+{e}^{-2t})dt=\frac{{e}^{2t}}{2}-\frac{{e}^{-2t}}{2}+C \ \ =\frac{1}{2}{(\frac{x}{e})}^{2x}-\frac{1}{2}{(\frac{e}{x})}^{2x}+C \end{gathered}$

On comparing I with $\frac{1}{\alpha }{(\frac{x}{e})}^{\beta x}-\frac{1}{\gamma }{(\frac{e}{x})}^{\delta x}+C$

$\begin{gathered} \alpha =2,\beta =2,\gamma =2,\delta =2 \ 	herefore \alpha +2\beta +3\gamma -4\delta =2+2	imes 2+3	imes 2-4	imes 2=4 \end{gathered}$

Solutions · 1