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$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x$ is equal to
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Verified Answer
The correct answer is:
$0$
Let $I=\int_{-\pi / 4}^{\pi / 4} x \tan \left(1+x^2\right) d x$
Here, $f(x)=x \tan \left(1+x^2\right)$
$f(-x)=-x \tan \left(1+x^2\right)=-f(x)$
$f(x)$ is an odd function.
$\therefore \quad I=0$
Here, $f(x)=x \tan \left(1+x^2\right)$
$f(-x)=-x \tan \left(1+x^2\right)=-f(x)$
$f(x)$ is an odd function.
$\therefore \quad I=0$
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