 In this video, we provide the solution to question number four for practice exam number two for math 1220 In which case we have to evaluate the improper integral from 0 to infinity of the function tangent inverse of x over 1 plus x squared dx So my first thoughts when I look at this is actually to do a u substitution because when you look at this You have tangent inverse of x the derivative of tangent inverse. Let's think about that for a second the derivative is Is actually going to equal 1 over 1 plus x squared dx So that's exactly what we have right here So I could rewrite this integral as simply though the tangent inverse becomes a u the dx over 1 plus x squared becomes a du That's fantastic, but then we need to adjust the bounds as well So if we take 0 and plug it into our tangent here, we're going to get our tangent of 0 which is likewise 0 like so but now as x goes to infinity what happens to our tangent well our tangent as x approaches infinity will approach its horizontal Ascentile which is pi halves. So actually this upper bound becomes pi halves in that situation And so by the u substitution, we've turned our improper integral actually into a proper integral We find an anti derivative of u which would be u squared over 2 we evaluate from 0 and pi halves When you plug in pi halves You'll get one half times pi Half squared When you plug in zero everything will just vanish so I can just get away with this One times pi squared of course is pi squared and then in the denominator You get a two square which is four times another two which is eight And so the correct answer would then be pi squared divided by eight and so we see that is choice E