An improper integral of a rational function

Definition. An improper integral $\int_0^{+\infty}f$ converges if $\lim_{R\to+\infty}\int_0^Rf$ exists and is finite.

Computation. For $R>0$, $\int_0^R\frac{dx}{1+x^2}=\arctan R-\arctan0=\arctan R\to\frac\pi2$ as $R\to+\infty$.

Answer. It converges, and
$$\int_0^{+\infty}\frac{dx}{1+x^2}=\frac\pi2.$$

Remark. The region under $\frac1{1+x^2}$ is infinitely long but has finite area $\frac\pi2$, because the curve decays like $\frac1{x^2}$, fast enough.