For α ,β ,γ ,δ ∈ mathbbN, if ∫ (( x e)^2x+( e x)^2x) log_exdx= 1 α ( x e)^β x- 1 γ ( e x)^δ x+C, where e=Σ _n=0^∞ 1 n! and C is constant of…

Given,

$$\int ({(\frac{x}{e})}^{2x}+{(\frac{e}{x})}^{2x})\mathrm{ln}xdx=\frac{1}{\alpha }{(\frac{x}{e})}^{\beta x}-\frac{1}{\gamma }{(\frac{e}{x})}^{\delta x}+C$$∫((ex​)2x+(xe​)2x)lnxdx=α1​(ex​)βx−γ1​(xe​)δx+C

Now let $\displaystyle I=\int ({(\frac{x}{e})}^{2x}+{(\frac{e}{x})}^{2x})\mathrm{ln}xdx$I=∫((ex​)2x+(xe​)2x)lnxdx

Now let ${(\frac{x}{e})}^{2x}=t$(ex​)2x=t

$$\Rightarrow 2x(\mathrm{ln}x-1)=\mathrm{ln}t$$⇒2x(lnx−1)=lnt

$$\Rightarrow \mathrm{ln}xdx=\frac{1}{2t}dt$$⇒lnxdx=2t1​dt

So, $\displaystyle I=\frac{1}{2}\int (t+\frac{1}{t})\frac{dt}{t}$I=21​∫(t+t1​)tdt​

$$\Rightarrow I=\frac{1}{2}\int (1+\frac{1}{{t}^{2}})dt$$⇒I=21​∫(1+t21​)dt

$$\Rightarrow I=\frac{1}{2}(t-\frac{1}{t})+c$$⇒I=21​(t−t1​)+c

$$\Rightarrow I=\frac{1}{2}({(\frac{x}{e})}^{2x}-{(\frac{e}{x})}^{2x})+C$$⇒I=21​((ex​)2x−(xe​)2x)+C

Now on comparing with $I=\frac{1}{\alpha }{(\frac{x}{e})}^{\beta x}-\frac{1}{\gamma }{(\frac{e}{x})}^{\delta x}+C$I=α1​(ex​)βx−γ1​(xe​)δx+C, we get $\alpha =2,\beta =2,\gamma =2,\delta =2$α=2,β=2,γ=2,δ=2

Hence, $\alpha +2\beta +3\gamma -4\delta =2+4+6-8=4$α+2β+3γ−4δ=2+4+6−8=4

Hence this is the required option.