JEE Main 2019MathematicsIndefinite IntegrationEasyMCQ

JEE Main 2019 — Indefinite Integration Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

If ∫dxx31+x623 =xfx1+x613+C, where C is a constant of integration, then the function fx is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B-12x3

Step-by-step explanation

We have, ∫dxx31+x623=xfx1+x613+C

Now, I=∫dxx71x6+123

Put, t=1x6+1⇒dt=-6x7dx

∴I=-16∫dtt23=-12t13+C, where C is the constant of integration=-121x6+113+C=-121+x613x2+C

∴fx=-12x3

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About this question

This is a previous-year question from JEE Main 2019, 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.