 In this video, we provide the solution to question number 10 for the practice exam number two for math 1220, in which case we have to evaluate the indefinite integral of the function x cube plus x plus one all over x squared plus one with respect to x. And so when I look at this, this is a rational function. I do notice that it's an improper rational function. Notice we have an x cubed on top, an x squared on the bottom, the numerator is larger degree than the denominator. So the very first thing I'm going to do is actually do long division of polynomials. So I'm going to take my x cubed plus x plus one here. I left the gap for the x squared term. Then we have an x squared plus one. So then we have to ask ourselves how many times is x squared divide into x cubed. And that really is just simplifying the ratio x cubed over x squared. That's equal to an x. And so I'm going to write that on the top of the line right there. I'm then going to take this polynomial and times everything by x, x squared times x is x cubed like we just were computing. We're also going to get one times x. So we get a plus x right here, but we're actually going to subtract this from above. So the x cubes are going to cancel x cubed minus x cubed. But I also get x minus x as well. So they actually cancel out again. And so then the next thing to do is to bring down the one. So we get a one right here. But then at this point one is actually smaller than x squared with regard to degree. So this is actually the remainder. So we have the quotient up here. We have the remainder down here. So then if I were to reevaluate what I have here, I guess I can just write it here. Our integral is actually equal to, I should say the function is equal to x plus one over x squared plus one dx like so. I'm going to make this parenthesis look a little bit bigger. And so in this situation, we have to find an anti derivative of x. We have to find an anti derivative of one over x squared plus one for which the anti derivative of x is going to be x squared over two, the anti derivative of one over x squared plus one. That's actually the that's the derivative of our tangent. So the anti derivative is going to be our tangent of x. If you didn't remember that that's one you ought to know but if you didn't remember that you actually could do a trig sub and reevaluate that one from scratch. I'm not going to go through the details of that because this is one of the standard ones that we can have memorized. Don't forget the plus C because we are looking for an indefinite integral and that then gives us the anti derivative there x squared over two plus our tangent of x plus a constant.