int_{-frac{pi}{4}}^{frac{pi}{4}} x tan left(1 | vedclass

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(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x 	an \left(1+x^2ight) d x$. Define the integrand as $f(x) = x 	an \left(1+x^2ight)$. To chec

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(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x 	an \left(1+x^2ight) d x$. Define the integrand as $f(x) = x 	an \left(1+x^2ight)$. To check if the function is odd or even,we evaluate $f(-x)$: $f(-x) = (-x) 	an \left(1+(-x)^2ight) = -x 	an \left(1+x^2ight) = -f(x)$. Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function. According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) d x = 0$. Therefore,$I = 0$.

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x$ is equal to

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