$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{ e } x d x=\frac{1}{\alpha}\left(\frac{ x }{ e }\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{ e }{ x }\right)^{\delta x }+ C ,$
Where $e =\sum \limits_{ n =0}^{\infty} \frac{1}{ n !}$ and $C$ is constant of integration, then $\alpha+2 \beta+3 \gamma-4 \delta$ is equal to:
$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _e x d x=\int\left[e^{2(x \ln x-x)}+e^{-2(x \ln x-x)}\right] \ln x d x$
$x \ln x-x=t$
$\ln x \cdot d x=d t$
$\int\left(e^{2 t}+e^{-2 t}\right) d t$
$\frac{e^{2 t}}{2}-\frac{e^{-2 t}}{2}+C$
$=\frac{1}{2}\left(\frac{x}{e}\right)^{2 x}-\frac{1}{2}\left(\frac{e}{x}\right)^{2 x}+C$
$\alpha=\beta=\gamma=\delta=2$
$\alpha+2 \beta+3 \gamma-4 \delta=4$
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