JEE Main 2023MathematicsIndefinite IntegrationHardMCQ

JEE Main 2023 — Indefinite Integration Question with Solution

JEE Main 2023 (10 Apr Shift 2)

Question

For α,β,γ,δ∈ℕ, if   ∫xe2x+ex2xlogexdx=1αxeβx-1γexδx+C, where e=∑n=0∞1n! and C is constant of integration, then α+2β+3γ-4δ is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B4

Step-by-step explanation

Given,

∫xe2x+ex2xlnxdx=1αxeβx-1γexδx+C

Now let I=∫xe2x+ex2xlnxdx

Now let xe2x=t

⇒2xlnx-1=lnt

⇒lnxdx=12tdt

So, I=12∫t+1tdtt

⇒I=12∫1+1t2dt

⇒I=12t-1t+c

⇒I=12xe2x-ex2x+C

Now on comparing with I=1αxeβx-1γexδx+C, we get α=2, β=2,γ=2,δ=2

Hence, α+2β+3γ-4δ=2+4+6-8=4

Hence this is the required option. 

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About this question

This is a previous-year question from JEE Main 2023, covering the Indefinite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.