2. $\int_{0}^{\infty }{\frac{dx}{1+x^{2}}}=\int_{0}^{1}{\frac{dx}{1+x^{ 2}}}+\int_{1}^{\infty }{\frac{dx}{1+x^{2}}},$ since the first piece is continuous in $[0,1]$ thus is integrable there (the integral exists, which means that it converges), now, for the second piece, we have that for $x\ge1$ it's $\frac{1}{1+x^{2}}<\frac{1}{x^{2}},$ since $\int_1^{\infty}\frac{dx}{x^2}<\infty,$ then so does $\int_{1}^{\infty }{\frac{dx}{1+x^{2}}},$ whereat the whole integral converges.