Mathematics · 2022
JEE Main · 26 June 2022, Shift 2 · Q28
The integral 24/(π) ∫_0^√ 2 ((2-x^2) d x)/((2+x^2) √(4+x^4)) is equal to \ \ \ \.
The integral $\displaystyle \frac{24}{\pi} \int_0^{\sqrt{2}} \frac{\left(2-x^2\right) d x}{\left(2+x^2\right) \sqrt{4+x^4}}$ is equal to \_\_\_\_.
Official answer
3
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.