The integral _1^e ( ( xe )^2x- ( ex )^x ) _ex\,dx is equal to :

Question

Accepted Answer

Correct answer: D. (d) JEEN Main 2019 I= ( ( xe )^2x- ( ex )^x) log_e xdx Let ( xe )^x=t ⇒x ln ( xe )= lnt ⇒x( lnx-1)= lnt On differentiating both sides w.r.t. x we get lnx . dx= dtt When x=e then t=1 and when x=1 then t= 1e I= (t^2- 1t) ∙ dtt= (t- 1t^2) dt =( t^22+ 1t)_ 1e^1=( 12+1)-( 12e^2+e)= 32-e- 12e^2

$The integral \int_{1}^{e} ((\frac{x}{e})^{2 x} - (\frac{e}{x})^{x}) lo g_{e} x d x is equal to :$

Solution

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