Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 27

int_-infty^infty 4/(16+x^2)dx=
Divide both numerator and denominator by 16.
int_-infty^infty (1/4)/(1+x^2/16)dx=int_-infty^infty (1/4 dx)/(1+(x/4)^2)=
Substitute u=x/4 => du=1/4 dx. Bounds of integration remain the same because dividing infinity (or minus infinity) by four gives infinity.
int_-infty^infty (du)/(1+u^2)=arctan u|_-infty^infty=lim_(u to infty)arctan u-lim_(u to -infty)arctan u=
pi/2-(-pi/2)=pi
As we can see the integral converges and its value is equal to pi.
The image below shows the graph of the function and area under it corresponding to the integral. As we can see the graph of the function approaches x-axis (function converges to zero) as x goes to +-infty. We can also see that this convergence seams pretty "fast" which means that the integral should converge which we've already shown by calculating it.