Mathematics · 2026
JEE Main · 4 April 2026, Shift 2 · Q20
The integral ∫_0^1 cot^-1(1+x+x^2) d x is equal to:
The integral $\displaystyle \int_0^1 \cot ^{-1}\left(1+x+x^2\right) d x$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 2 \tan ^{-1} 2-\frac{1}{2} \log _e\left(\frac{5}{4}\right)-\frac{\pi}{2}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.