JEE Main 2022MathematicsIndefinite IntegrationMediumMCQ

JEE Main 2022 — Indefinite Integration Question with Solution

JEE Main 2022 (26 Jun Shift 2)

Question

If ∫1x1-x1+xdx=gx+c,g1=0, then g12 is equal to

Choose an option

Show full solutionCorrect option: A
Correct answer
Aloge3-13+1+π3

Step-by-step explanation

Given,

∫1x1-x1+xdx=gx+c

Put x=cos2θ

dx=-2sin2θ·dθ

=∫1cos2θtanθ-4sinθ·cosθdθ

=∫1cos2θ-4sin2θdθ

=-2∫1-cos2θcos2θdθ

=-22lnsec2θ-tan2θ+2θ+c

=lnsec2θ-tan2θ+2θ+c

=ln1-sin2θcos2θ+cos-1x+c

=ln1−1−x2x+cos−1x+c⏟gx

∴  g1=0⇒c=0

So, gx=ln1-1-x2x+cos-1x

g12=ln2-3+π3

g12=ln3-13+1+π3

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About this question

This is a previous-year question from JEE Main 2022, 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.