JEE Main 2018MathematicsIndefinite IntegrationHardMCQ

JEE Main 2018 — Indefinite Integration Question with Solution

JEE Main 2018 (15 Apr)

Question

If fx-4x+2=2x+1, x∈R-1,-2, then∫fxdx is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B-12ln1-x-3x+C

Step-by-step explanation

We have, fx-4x+2=2x+1

Let x-4x+2=X⇒xX+2X=x-4⇒xX-1=-4-2X⇒x=22+X1-X

Putting x=22+X1-X in the given functional equation, we get 

⇒fX=2×22+X1-X+1=8+4X+1-X1-X=3X+91-X=3X+31-X

Replacing X→x, we get

fx=3x+31-x

⇒∫fxdx=3∫x+31-xdx=3∫4-1-x1-xdx=3∫41-xdx-∫dx

=-12ln1-x-3x+C, where C is the constant of integration.

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About this question

This is a previous-year question from JEE Main 2018, 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.