Mathematics · 2021
JEE Main · 18 March 2021, Shift 1 · Q73
The integral ∫ ((2 x-1) cos √((2 x-1)^2+5))/(√(4 x^2-4 x+6)) d x is equal to: (where c is a constant of integration)
The integral $\displaystyle \int \frac{(2 x-1) \cos \sqrt{(2 x-1)^2+5}}{\sqrt{4 x^2-4 x+6}} \mathrm{~d} x$ is equal to : (where c is a constant of integration)
Official answer
(2)
$\displaystyle \frac{1}{2} \sin \sqrt{(2 x-1)^2+5}+c$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.