 In this video, we're gonna discuss the solution to question 11 from the practice midterm exam for calculus two math 12-20. And so we wanna evaluate the indefinite integral where we have a function x cubed plus x plus one over x squared plus one. And we wanna integrate with respect to x here. So the first thing I notice here is that we have an improper fraction. We have an x cubed over x. And so as such, we're gonna wanna do some polynomial division. We could try to weasel our way around a couple things. One of the most straightforward things is just to do long division, right? If you divide the top x cubed plus x plus one, we divide that by x squared plus one. We're gonna first ask ourselves how many times does x squared go into x cubed? Look at the leading terms there. x cubed over x squared, that gives you an x. We're gonna record that on the top right here. Next, we're gonna take x times our divisor x squared plus one. This gives us x cubed plus x. And we're gonna record that below x cubed plus x. Like so, and we're gonna subtract that from above. You're gonna notice here that the x squared, the x cubed cancel and then also the x cancels as well. There's really not much left. You just end up with a one right here. One is too small to work with here. And so this is actually gonna be a remainder. And so what we can do is we can then substitute our function with now, we're going to get the polynomial x plus one over x squared plus one, dx. So that polynomial division is gonna be super helpful for us on this example. Now in this example, we could have actually got weaseled away from the polynomial division because it actually turned out so simply. We actually could have done something like the following. If you take x cubed plus x over x squared plus one and you break off the one over here, the idea is x cubed plus x is just x times x squared plus one for which that cancels, give me the same thing. So this one could have, we could have avoided long division doing some type of short division. But what I'm trying to say is you should just plan for long division if the fraction isn't proper. If it's a proper fraction, it's completely unnecessary. So now the question is, what do we do here? So do we, so the entire derivative of x is gonna be pretty straightforward. But this one right over here, we have this x squared plus one on the bottom. It's an irreducible quadratic polynomial. So, I mean, notice this is a proper fraction. In fact, the denominator doesn't factor at all. This is already a partial fraction decomposition. Although you might need to do a partial fraction decomposition on question number 11, not necessary here. Instead, how do we do this one? Well, two approaches, you could do a u substitution because you have this square root of x squared plus one squared on the bottom. You could try the u substitute, not u substitution, trig substitution, x equals tangent theta. That would be perfectly acceptable in which case then you get dx equals secant squared theta d theta. And notice the square root of x squared plus one would equal secant theta as well. So the anti-derivative of x is gonna be x squared over two. And then you have to integrate. The dx, remember, becomes a secant squared theta d theta. Since the square root of x squared plus one equals secant, the bottom would then become a secant squared theta. In which case those cancel out. You have this x squared over two plus the integral of d theta. Well, that's pretty nice. It just becomes theta plus a constant. Don't forget the plus constant. I will look for that there. And so then coming back to original substitution, solving for theta, theta equals tangent inverse of x. And so our final result would then be x squared over two plus arc tangent of x plus a constant. And this would be our correct result. In order to get full credit, if we do any type of substitutions, you need to switch your variables back into x. You need all x's, no thetas or u's or anything else. That's okay in the interim, but for the final answer, it needs back in the original variable x. I also wanna mention that many of you might have noticed earlier that if you take the anti-derivative of one over one plus x squared dx, you might already know that this anti-derivative is tangent inverse of x and you could have jumped the whole trig substitution part and gone straight to there. I'm perfectly fine with that. This is a very common anti-derivative and therefore recognizing it without any trig sub is perfectly appropriate. But if you didn't recognize it, be aware that the trig substitution would help you out here. Now, this one was a partial fraction type problem, although we didn't actually do any partial fractions. We used long division and we sort of did a trig sub. It depends whether you did it or not. Those are unnecessary parts for this type of questions from 7.4, but one of the most important parts is actually doing the partial fraction decomposition. And so just because you didn't see the PFD on this problem, on the practice test, doesn't mean it won't show up on the actual test. You're gonna make sure you're familiar with factorization techniques, the templates and all of that stuff as well.