$\lim_{r\to\infty} \int_0^\infty rx/(exp(x) + r^2x)dx = 0$

You can find a dominating function by differentiating the integrand with respect to $n$. If not clear, here are more details.

Fix $x>0$. For $t\in [1,\infty)$, define $f_t (x) = \frac{t x} {e^x + t^2 x}$.

$$\frac{d}{dt} \frac{1}{f_t (x)} = \frac{d}{dt} (\frac{e^x + t^2x}{tx}) = -\frac{1}{t^2} \frac{e^x}{x}+1.$$

This derivative is increasing in $t$, is negative at $t=1$ ($e^x>x$ for $x>0$) and tends to $1$ as $t\to\infty$.

Therefore the unique minimum over $t \ge 1$ is attained when $t^2 = e^{x}/x$. As a result, for all $t\ge 1$ we have

$$f_t (x) \le f_{e^{x/2}/\sqrt{x}}(x) = \frac {\frac{e^{x/2}} { \sqrt{x} }x} {e^x + e^x} =\frac{1}{2} \sqrt{x} e^{-x/2}.$$

This upper bound clearly holds for all $t\in{\mathbb N}$. Furthermore the function on the righthand side is integrable. Therefore we can apply the dominated convergence theorem with it as the dominating function.