Notice:
$$f(x) = x(\arctan x)'=x \cdot \frac{1}{1+x^2}$$
This is an **odd** function (i.e. $f(x) = -f(-x)$), and thus the integration result is $0$.
**EDIT per comment**
The integration does not exist unless we consider Cauchy principle value:
$$\int_{-\infty}^{+\infty}f(x) \,dx =: \lim_{R \rightarrow +\infty}\int_{-R}^Rf(x)\,dx $$
And in our case, the Cauchy PV is $0$.