 Let's take a look at the solution to question number two for our practice exam for the midterm in calculus 2 math 12 20 here And so in this in this one right here We have an improper integral integrated from 0 to infinity arc tangent of x over 1 plus x squared dx Well that art tangent on top and the 1 plus x at first at the beginning it looks impenetrable But really this actually leads itself to a very nice u substitution if we take u to be arc tangent of x Then it's derivative du be dx over 1 plus x squared Which actually works out really great here because this is our du Right front and center there, which really will simplify the integral just to be u du, which I can handle that I really like that a lot. Let's switch the boundaries as well As x ranges from 0 to infinity Infinity what happens to you well arc tangent of 0 isn't so bad That's just gonna be a 0 itself But then what happens here when we take arc tangent of infinity what we're talking about is a limit, right? So it's x approaches infinity what happens to you well as x approaches infinity arc tangent will approach Pi haves its horizontal asymptote and so that's actually what we record over here pi haves and So notice by taking the limit right now When we do the u substitution our improper integral actually switches to a proper one, right? In which case then we have to find the anti-derivative Which will be u squared over 2 as you go from 0 to pi haves when you plug in 0 everything will disappear So we get one half times pi over 2 Squared and so in the end that would simplify to be pi squared over 8 which gives us option e right here