JEE Main 2015MathematicsDefinite IntegrationMediumMCQ

JEE Main 2015 — Definite Integration Question with Solution

JEE Main 2015 (10 Apr Online)

Question

For x>0, let fx=∫1xlog⁡t1+t dt.  Then fx+f1x is equal to 

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Show full solutionCorrect option: A
Correct answer
A12 log⁡x2

Step-by-step explanation

fx= ∫1xlog⁡t1+t dt ⇒fx=∫1xlog⁡z(1+z)dz
And f1x= ∫11xlog⁡t1+t dt
Put t=1z
dt= -1z2 dz
fx= ∫1xlog⁡zz21+ 1z ⋅dz
fx= ∫1xlog⁡zz(1+z)  dz
fx+f1x= ∫1xlog⁡z 11+z+1z(1+z) dz
= ∫1x1zlog⁡z dz
Put log⁡z=P⇒1z dz=dP
∫1xP⋅dP=P221x=12 log2⁡z1x= log⁡x22

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About this question

This is a previous-year question from JEE Main 2015, covering the Definite 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.