For , if [ ( x e ) 2 x + ( e x ) 2 x ] log e x ? d x = 1 ( x e ) x - 1 ( e x ) x + C , where e = n = 0 1 n and C is constant of integration, then + 2 + 3 - 4 is equal to [2023]

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Solution

Q.
$For α, β, γ, δ \in ℕ, if \int [{(\frac{x}{e})}^{2 x} + {(\frac{e}{x})}^{2 x}] \log_{e} x d x$ $= \frac{1}{α} {(\frac{x}{e})}^{β x} - \frac{1}{γ} {(\frac{e}{x})}^{δ x} + C, where e = \sum_{n = 0}^{\infty} \frac{1}{n!}$ $and C is constant of integration, then α + 2 β + 3 γ - 4 δ is equal to$ [2023]

1 1

2 - 8

3 4

4 - 4

Ans.
(3)

$We have,$ $\int [{(\frac{x}{e})}^{2 x} + {(\frac{e}{x})}^{2 x}] \log_{e} x d x$

$Putting {(\frac{x}{e})}^{2 x} = t$ $\Rightarrow 2 x (\log x - 1) = \log t$

$\Rightarrow [2 (\log x - 1) + 2] d x = \frac{1}{t} d t \Rightarrow 2 \log x d x = \frac{1}{t} d t,$ we get

$\int (t + \frac{1}{t}) \frac{1}{2 t} d t = \frac{1}{2} \int d t + \frac{1}{2} \int \frac{1}{t^{2}} d t$

$= \frac{1}{2} t - \frac{1}{2 t} + c$ $= \frac{1}{2} [{(\frac{x}{e})}^{2 x} - {(\frac{e}{x})}^{2 x}] + c$

$Comparing with \frac{1}{α} {(\frac{x}{e})}^{β x} - \frac{1}{γ} {(\frac{e}{x})}^{δ x} + c$

$We get, α = 2, β = 2, γ = 2, δ = 2$

$Now, α + 2 β + 3 γ - 4 δ = 2 + 4 + 6 - 8 = 4$

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