The function is symmetrical about the $y$-axis. Break up the integral into two parts, $-\infty$ to $0$ and $0$ to $\infty$. By symmetry, we need only deal with $\int_0^\infty \frac{x^\alpha}{(1+x^2)^m}dx$
We are not told anything about $\alpha$, or indeed about $m$. In particular, $\alpha$ could be negative, which would make the integrand blow up near $x=0$. And of course integrals over an infinite interval raise issues of convergence.
Let's break up the region of integration into two parts, say from $0$ to $1$ and from $1$ to $\infty$. (Instead of $1$ we could use any positive number.)
You probably know that $\int_0^1 x^\alpha dx$ converges if $\alpha>-1$ and diverges if $\alpha \le -1$.
Informally, near $0$, the function $x^\alpha/(1+x^2)^m$ "behaves like" $x^\alpha$, since if $x$ is near $0$, $(1+x^2)^m$ is almost $1$.
So informally, the convergence behaviour of the integral should be the same as the (known) convergence behaviour of the integral of $x^\alpha$.
Semi-formally, we have the following result. Suppose that $f(x)$ and $g(x)$ are positive functions, which are well-behaved everywhere on $(0,b]$ but may blow up as $x \to 0$. If $\lim_{x\to 0}\frac{f(x)}{g(x)}=1$, then $\int_0^bf(x)dx$ converges if and only if $\int_0^b g(x)dx$ converges.
For our integral from $0$ to $1$, let $f(x)$ be our function, and let $g(x)$, the comparison function, be $x^\alpha$. It is very easy to show that the limit of the ratio $f(x)/g(x)$ as $x \to 0$ is $1$. So our integral from $0$ to $1$ converges if and only if $\int_0^1 x^\alpha dx$ converges, and we know everything about that.
Finally we look at the integral from $1$ to infinity. It is useful to deal separately with positive and negative $m$. In fact, from the notation, I expect that implicitly or explicitly $m$ is supposed to be positive.
For the integral to exist, the function must approach $0$ as $x\to \infty$. So we must have $2m>\alpha$. But that's not enough: the function must go down kind of fast.
Informally, if $2m>\alpha$, then for large $x$ our function behaves like $g(x)$, where $g(x)= 1/x^{2m-\alpha}$. (Note that this is not the same $g(x)$ as the one we used for the integral from $0$ to $1$.) We know that $\int_1^\infty 1/x^k dx$ converges if and only if $k>1$.
If our function is $f(x)$, it is easy to show that $\lim_{x\to\infty}f(x)/g(x)=1$.
So we can conclude that our integral from $1$ to $\infty$ converges if and only if $2m-\alpha>1$.
Finally, for our full integral to converge, both parts ($0$ to $1$, $1$ to $\infty$) must converge.
We conclude that our integral converges iff the two conditions $\alpha>-1$ and
$2m-\alpha>1$ both hold.
This can be written more nicely as $-1 < \alpha <2m-1$