Mathematics · 2019
JEE Main · 9 April 2019, Shift 2 · Q75
The value of the integral ∫_0^1 x cot^-1(1-x^2+x^4) d x is:
The value of the integral
$$\int_0^1 x \cot ^{-1}\left(1-x^2+x^4\right) \mathrm{d} x \text { is : }
$$
Official answer
(1)
$\displaystyle \frac{\pi}{4}-\frac{1}{2} \log _e 2$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.