For α , β , γ , δ ∈ ℕ , if    ∫ x e 2 x + e x 2 x log e x d…

Given,

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

Now let $I=\int \left({\left(\frac{x}{e}\right)}^{2x}+{\left(\frac{e}{x}\right)}^{2x}\right)\ln xdx$

Now let ${\left(\frac{x}{e}\right)}^{2x}=t$

$\Rightarrow 2x\left(\ln x-1\right)=\ln t$

$\Rightarrow \ln xdx=\frac{1}{2t}dt$

So, $I=\frac{1}{2}\int \left(t+\frac{1}{t}\right)\frac{dt}{t}$

$\Rightarrow I=\frac{1}{2}\int \left(1+\frac{1}{t^2}\right)dt$

$\Rightarrow I=\frac{1}{2}\left(t-\frac{1}{t}\right)+c$

$\Rightarrow I=\frac{1}{2}\left({\left(\frac{x}{e}\right)}^{2x}-{\left(\frac{e}{x}\right)}^{2x}\right)+C$

Now on comparing with $I=\frac{1}{\alpha }{\left(\frac{x}{e}\right)}^{\beta x}-\frac{1}{\gamma }{\left(\frac{e}{x}\right)}^{\delta x}+C$, we get $\alpha =2, \beta =2,\gamma =2,\delta =2$

Hence, $\alpha +2\beta +3\gamma -4\delta =2+4+6-8=4$

Hence this is the required option.