JEE Main 2020MathematicsIndefinite IntegrationMediumMCQ

JEE Main 2020 — Indefinite Integration Question with Solution

JEE Main 2020 (03 Sep Shift 2)

Question

 If ∫sin-1x1+xdx=Axtan-1x+Bx+C, where C is a constant of integration, then the ordered pair Ax, Bx can be :

Choose an option

Show full solutionCorrect option: D
Correct answer
Dx+1, -x

Step-by-step explanation

I=∫sin-1x1+xdx

=∫tan-1x⏟I1⏟IIdx

=xtan-1x-∫11+x.12x.xdx+C

=xtan-1x-12∫t.2t.dt1+t2+C (putting x=t2⇒dx=2tdt)

=xtan-1x-∫t21+t2dt+C

=xtan-1x-t+tan-1t+C

=xtan-1x-x+tan-1x+C

=x+1tan-1x-x+C

Comparing with Axtan-1x+Bx+C, we get

Ax, Bx=x+1, -x

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About this question

This is a previous-year question from JEE Main 2020, 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.