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Problem 30 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{x^3 - 2x^2 + 2x - 5}{x^4 + 4x^2 + 3}\ dx $

Answer

$$
\frac{1}{12}\left[3 \ln \left(x^{2}+1\right)+3 \ln \left(x^{2}+3\right)-18 \tan ^{-1} x-2 \sqrt{3} \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C\right]
$$

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Video Transcript

Okay, So this is a quite challenging problem that requires you to know quite a lot regarding, um, integration. And it takes a lot of algebra to go through it. So I'm gonna go through the big concepts that you want. Thio really pay attention to. I cannot go through each off the steps because that's going to take more than 20 minutes for me to explain everything. But hopefully it will help you solve this problem by yourself. It's going to be very good practice. Okay, so let's start with this expression. You see that the numerator is a cubic polynomial. The denominator is a four degree polynomial. So at this point already, you should all you should start by thinking, Okay, Can you substitution work? Because if I do use substitution and the denominator will produce a cubic and it might cancel with the top. But the thing is, when I do that four x cubed plus eight x, it doesn't really seem to appear on the numerator, so that doesn't work. So the next thing you might want to consider is off course. It's not one of the easiest things that you could do, but it is partial fraction the composition. Okay. And in this case, what makes things a little bit more complicated is the fact that the denominator is a quart IQ and you end up having irreducible, quadratic ksaz there factors which makes the decomposition a little bit tough to solve. So I'm going to go through the idea of it. Okay, so let's start with that portion first is considered the decomposition. So in the decomposition step, what you're going to basically say is let me call this portion Delta. First, you're going to factor Delta. You will end up having if you notice that this is a special form of the quadratic. It is a square plus for a plus three, so you can factor that quite straightforwardly calling it X squared plus one times x court plus three. So here you see that X Square plus one and Expert Square plus three are not reducible by linear factors anymore. So these air called the irreducible quadratic. Okay, So what that means is when I do have partial fraction decomposition. Uh, this is what what it's gonna look like X squared plus one. The numerator is going to be a linear expression explicit E. In the other part, X squared plus three, you will have another linear expression, so you have to solve a, B, C and D. It's going to be a system of equations with four variables, which takes a very long time to solve by hand. So it's one of the reasons why um, this problem could be a little bit tough. Okay, because the longer time that you spend doing algebra, the more likely that you make a small mistake, which you cannot really catch until you check the answers at the end. So makes it a tough one. Anyhow, using this basically where you're going to say is this is equal to X cubed minus two x squared plus two x minus five. Basically, the numerator off, the integration, the denominator. I'm just going to write Delta. What you're going to do is to multiply Delta on both sides so you will get X cubed plus three times a X plus B x squared, plus one times see explicit e and the numerator heart. Let's call it square That ends up being the numerator. Okay, Now there are two paths that most people take whenever they do partial fractions. One type of people will actually expand the left side, simplify it, and then they compare the the coefficient off each of the terms that you have on the left and the right. So, for example, on the left side, you will have a Oh, that should not be acute. That should be a square. There is no you will notice that all X cubed in ah si X cubed shows up. So a plus c has to be one. You can also see that three B and D has to be a negative five because those are the constant terms. So just like that, you're you might be able to calculate, um, the coefficients compare them and you end up having a system of equations. That's one path. Another path that I usually like to take is to start from the equation that you have right here and start plugging in numbers that usually works extremely well when you have a linear factor instead of a quadratic. Because if you have linear factors by plugging in the number that you see right next to the variable or the opposite of it, you can actually make it equal to zero, simplifying the expression significantly. So, um, in this case, it's a quadratic, so it's a little bit tougher, but I will demonstrate how I did it. So you can always start with easy numbers like X equals to zero. So when you plug in X equals to zero, you can see that the first term is three NMB, and the next time you will get one times D. So three B plus D is the left term. On the right side, you see that the first three terms has a zero in a zero in it, so you will only get a negative five. So that's an expression that you're going to get. I'm not going to go through the precise calculations, but I did. X equals to one X equals to negative. One actually goes to Tuas well, and then I get a system of equation that looks like this for a plus four B plus two C plus two d equals to negative four. The next term is negative for a plus four B minus to C plus two D, Thank you. There's 10 and the last one is 14 a plus seven b plus 10 C was five d equals to negative one. Okay, Now, what's nice, if you can see here? Is that the second and the third equation? You can see that the A's and the seas can cancel each other While B and D will remain so by using that fact, the first equation and the last two equations is just gonna be a system of equations of bees and these so you can figure out what B and d are right away, which makes the calculation just a little bit easier. Okay, and of course, you would still have to use the fourth equation, which is kind of ugly, but you will end up having I already wrote it down A is one half as well as see B is negative. Three halves and D is negative one half. Okay, now, this took me about 10 minutes just to solve these values. So again, I'm not gonna have enough time to go through it in detail, But that's the big idea of what actually is going on. Mhm. Let me move this aside. So we're done with this part. So what does that entail? Well, I was originally trying to solve it in this form. So let's figure out what this expression ends up being through this process, I was able to break this down into We'll go. Let me get my notes. Found that it is going to look like the integration off 1/2 X over X squared plus one minus three over to one over X squared, plus one plus one half X over X cubed plus three. I mean, X square plus three minus one half, one over X squared plus three d x. Okay, so it looks longer. But each of these integrations are doable. Okay? It's not easy, but its doable. So another big picture that we need to see here Two things they're going to happen. Hopefully, you are familiar with some of these expressions like this one. If you instantly remember that this is a tangent inverse, that's going to be a huge advantage of you. Okay, A similar thing happens with this one, but, um, not even I memorized these kind of formula all the time. It's actually something that I have to derive by myself. And if you look at the beginning off the formula sheet in calculus books, you usually confined it there. But it's very unlikely that most people have have memorized that. So I will actually go through how to calculate this. Integral. And I would assume that you would also be able to do it for this one. But it's more likely that you actually memorized it. Okay, Now these two can be solved exactly the same way. It's just a straightforward use institution. So I will go through one of them just to demonstrate how you would do it. Okay. All right. So let's start with the easy one. The red one, the use substitution This portion if I let u equals two x squared plus one, I know that. Do you equals to two x d. X. So x DX is do you over to Okay, now, if I substitute that in the integration off X over X squared plus one, I'm going to write the d X on the top just to make it look a little bit easy to follow. This portion is Do you divided by two. This portion is you. So, by substitution, I was able to get one half integration off. Do you over you or one over you do you if you prefer that notation. But we all know that this is just a natural log off absolute value of you and you is equal to X squared plus one. So what's nice here is that we actually don't need the absolute value, because export plus one is always positive. So you can rewrite it as the natural log off expert plus one in parentheses. You don't have to call it absolute value, Okay? And of course, we have a plus C because this is an indefinite in the role, but I will omit it for, um, just for simplification. Purpose. All right, moving on. You do the same thing with this expression, so I'm not going to go through that much in detail. Knicks many move this. Okay, now, this one, the steps that I am going to take with the other one will actually show you why the anti derivative off This will be a tangent in verse, and you basically use tricks substitution to do so and, um, hopefully you'll be able to follow it because it is quite intense. Okay, so let's take a look at the integration off one over X squared, plus three. Now when you learn tricks substitution, you kind of learn the trick that trigonometry starts out as a study of right triangles and similar triangles. So things that are related with squares usually go very well with trigonometry. Another fact that you could remember is that sign square plus cold science squared equals to one Qianjin square plus one is equal to seek and squared etcetera. So the Pythagorean identity off the Trigana metric expressions are also quite useful to So that's basically the idea that we're going to use. So here the first step that we can take is to factor out of three and make it look like this. This is X divided by Route three squared because if I take out of three and if I take if I pull out a square out of it, it will be a square root, okay? And I pulled out a three from three. So that's just the one. So it looks like something squared, plus one. So if that something is a tangent, Qianjin squared plus one is seeking squared, which makes it a little bit easier to deal with. And something really nice is gonna happen next So that's why this technique is awesome And then we will do the trick Substitution Qianjin Fada is equal to x over three So if we do tangent squared plus one, the denominator becomes seek and squared. Now this is the part that's amazing. If I take the derivative with respect to X on both sides it turns out what's the derivative of tangent It's seek and squared so you will get c can't squared Fada is d X over route three So three seconds squared is what the X is equal thio Okay, so when I do the trick substitution, I'm gonna be able to see that the denominator is seeking squared. The X also has a Seacon square, so they cancel each other out. So this expression makes the integral look like Route 3/3 the integration off one di fada much, much easier to deal with. So this is, of course, Route 3/3 data. So my last step is, Well, what's data in terms of X? We already have this expression from here we can deduce that tangent oven angle is X divided by Route three. So that angle must be arc tangent of X over three. So 3/3. I don't have space. I'm just going to say tangent in verse. Off data. Okay, our X over there. Enough data. There you go. So the general form, I think it has a in it where that's the square root of a and stuff. I cannot remember exactly what it was, but this is basically the step you take in order to prove what it looks like. Okay. All right. So, going through all of these steps, I believe in you. You're going to be able to solve all of these integration When you move to these guys over here and the rest, I am just going to write down the answer what it would look like. So these two involves a natural log of X squared, plus something these two would involve. The Arc Tangent Hey, as the integral. And let's not forget that there's going to be a plus c at the end because this is an indefinite integral and, oh, I forgot to write down the solution, but it is going to look like let's see if I can actually write it here one half off one half of that will be 1/4 natural log of X square plus one. Let me write this one over here next. Because I already know what it should look like. There should be another 1/4 natural log of X squared plus three. Using the law of logarithms. You can combine these two if you want to. Just so you know. But I won't. OK. Minus minus 3/2. That's just tangent in verse. And the last one. It's negative. One half times this number. So negative. Route 3/3, not 36 See, it's such a long problem. I didn't even I wasn't even 100% prepared to write the solution so smoothly. Okay, so that is going to be tangent in verse X over three. Clause C. Right. So, numerically, it should be fine. At least concept wise. This is exactly what you're supposed to solve. This problem now. Good luck to you guys.