Mathematics · 2023
JEE Main · 6 April 2023, Shift 1 · Q11
Let I(x)=∫ (x^2(x sec^2 x+ tan x))/((x tan x+1)^2) d x. If I(0)=0, then I((π)/4) is equal to
Let $\displaystyle I(x)=\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x$. If $\displaystyle I(0)=0$, then $\displaystyle I\left(\frac{\pi}{4}\right)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \log _e \frac{(\pi+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.