Mathematics · 2024
JEE Main · 27 January 2024, Shift 2 · Q11
The integral ∫ ((x^8-x^2) d x)/((x^12+3 x^6+1) tan^-1(x^3+1/x^3)) is equal to:
The integral $\displaystyle \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\dfrac{1}{x^3}\right)}$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1 / 3}+C$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.