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]

Question

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

Smart Achievers · Accepted Answer

(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

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

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

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

$We get, α = 2, β = 2, γ = 2, δ = 2$

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

Solution

1 1

2 - 8

3 4

4 - 4

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Solution

Q. For α,β,γ,δ∈ℕ, if ∫[(xe)2x+(ex)2x]logex dx=1α(xe)βx-1γ(ex)δx+C, where e=∑n=0∞1n!and C is constant of integration, then α+2β+3γ-4δ is equal to [2023]

1 1

2 - 8

3 4

4 - 4

Ans. (3) We have, ∫[(xe)2x+(ex)2x]logex dx Putting (xe)2x=t ⇒2x(logx-1)=logt ⇒[2(logx-1)+2] dx=1tdt ⇒2logx dx=1tdt, we get ∫(t+1t)12t dt=12∫dt+12∫1t2 dt =12t-12t+c=12[(xe)2x-(ex)2x]+c Comparing with 1α(xe)βx-1γ(ex)δx+c We get, α=2, β=2, γ=2, δ=2 Now, α+2β+3γ-4δ= 2+4+6-8=4

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