Indefinite integral of an odd function

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In one of our classes the teaching assistant argued that $\int_\mathbb{R}ye^{-\frac{y^2}{2}}$ is an odd function and hence the integral is $0$.

But this argument doesn't hold in the case of $\int x \frac{1}{1+x^2}$.

In general when dealing with indefinite integrals I'm not allowed to use the argument odd-function, right?. I have to compute it by hand. So for the first example:

$\int_\mathbb{R}ye^{-\frac{y^2}{2}} = [-e^{-\frac{y^2}{2}}]|_{-\infty}^{+\infty} = 0 - 0 = 0$ would have been the right argument to say that the integral is $0$, right?

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If you can determine in advance that the integral over $\mathbb{R}$ is convergent, then you can say without further computation that it must be zero (under the assumption that the integrand is odd, of course).