Show X_n converge in probability if and only if $\lim\limits_{n \to \infty} \int_{-\infty}^{\infty}{\frac{x^2}{1+x^2} dF_n(x)} = 0.$

The condition $$ \lim\limits_{n \to \infty} \int_{-\infty}^{\infty}{\frac{x^2}{1+x^2} dF_n(x)} = 0$$ can be rephrased as $$ \lim\limits_{n \to \infty} \mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]= 0.$$

Note that for each positive $\varepsilon$, $$ \mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]\geqslant \mathbb E\left[\frac{X_n^2}{1+X_n^2}\mathbf{1}_{\{X_n^2>\varepsilon\}}\right]\geqslant \frac{\varepsilon}{1+\varepsilon}\mathbb P\left(X_n^2>\varepsilon\right) $$ since the map $t\mapsto t/(1+t)$ is increasing.

Moreover, $$ \mathbb E\left[\frac{X_n^2}{1+X_n^2}\right] =\mathbb E\left[\underbrace{\frac{X_n^2}{1+X_n^2}}_{\leqslant 1}\mathbf{1}_{\{X_n^2>\varepsilon\}}\right]+\mathbb E\left[\frac{X_n^2}{1+X_n^2}\mathbf{1}_{\{X_n^2\leqslant \varepsilon\}}\right] \leqslant \mathbb P\left(X_n^2>\varepsilon\right)+\varepsilon. $$ We thus got $$ \frac{\varepsilon}{1+\varepsilon}\mathbb P\left(X_n^2>\varepsilon\right)\leqslant \mathbb E\left[\frac{X_n^2}{1+X_n^2}\right]\leqslant \mathbb P\left(X_n^2>\varepsilon\right)+\varepsilon $$.