Pretty cool. Evaluate each integral.
int(0--> inf) [1/(1+x^2)^2]dx
int(0--> inf) [x^2/(1+x^2)^2]dx
Good solution - it's the way he did it. I just went directly to x = tan(u) in both cases and they led to integrals of sin^2(u) and cos^2(u) from 0 ---> pi/2. Try this and you will see this elegant solution. This tells us that they have to yield the same result since the area under both is the same from the symmetry between 0 and pi/2. It's not so much the actual value but the fact that they are the same is the most interesting. Look at the equations and their graphs - didn't see that one coming. Pretty cool.