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$\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=$
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Verified Answer
The correct answer is:
$\frac{\pi^2}{32}$
Put $t=\tan ^{-1} x \Rightarrow d t=\frac{1}{1+x^2} d x$, then
$\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=\int_0^{\pi / 4} t d t=\left[\frac{t^2}{2}\right]_0^{z / 4}=\frac{\pi^2}{32}$
$\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=\int_0^{\pi / 4} t d t=\left[\frac{t^2}{2}\right]_0^{z / 4}=\frac{\pi^2}{32}$
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