 Welcome back everyone. Let's take another look at an antiderivative of a rational function for which there's going to be irreducible quadratic denominator, right? With this one, you'll notice that if you ever have a quadratic polynomial ax squared plus bx plus c, we can try to factor it using, you know, multiply the a and the b together and look for factors to add up to be the a and the c together to get factors of b there. So here four times four times three gives us a 12. If you try to find factors of 12 to add to negative 4, it's not going to work very well here. And if you look at the quadratic formula, particularly at this discriminant, this b squared minus 4ac, in this situation you notice that you end up with this negative 4 squared which is 16 minus 4 times 4 times 3. You get 16 minus 48, which is negative 24. That's a negative number. And so what this tells us is that there's going to be no real roots, no real roots for this denominator, right? And so the denominator is already irreducible. And so we can't, unless we're going to use complex numbers, we can't break this up any more than we can already have in front of us. So how do we proceed from something like this? Well, the first thing to notice is that this thing is, even though there's an irreducible quadratic on the bottom, it is an improper fraction, right? Looking at the leading terms, the leading terms, you have 4x squared over 4x squared. That's the same power. We can reduce this down a little bit. We want a proper fraction. Now, that could be done using long division. So you have like 4x squared minus 4x plus 3 and divide that by 4. You get the idea. Well, I wrote them in the wrong spot there. Sorry, this should be a negative 3x plus 2. And then the other one goes out here. We could do that. But this turns out this one has a nice little slick move we could do here. So notice that the denominator is a 4x squared minus 4x plus 3. I'm going to break up into two pieces, okay? 4x squared minus 4x plus 3. The 4x squared kind of looks like it's already good to go, right? And then over here, we have a 9 is 3x plus 2. Now, what I need, what I need is I want a negative 4x right here and a plus 3. So that way this thing just cancels to be a one. But if you're going to do that, you have to do the opposite. If you subtract 4x, you have to add 4x to make sure things are still equal. If you add 3, you have to subtract 3. And so when you go about doing that, this thing will simplify to be 1 plus you get 4x minus 3x. That's just an x. And then you get negative 3 plus 2, which is a minus 1. This sits above our denominator 4x squared minus 4x plus 3. And so using this right here, right, we're going to integrate this thing. The x, integrate this thing. The anti derivative of 1 is just going to be x. So that's useful here. But then the second part, how do you deal with that, the integral of x minus 1 over 4x squared minus 4x plus 3 dx? Well, this is a situation where we would like to start off with a u substitution, take u to be the denominator 4x squared minus 4x plus 3, for which then du would equal 8x minus 4 du. Like so. Right? We get this 8x minus 4, in which case, if we were two times the whole thing by 8, get a 1 eighth right here, this would distribute to give me an 8x minus 8, which is almost what we want. We want an 8x minus 4. So actually the 8, we're going to break up into be a negative 4 and a negative 4. So breaking this up a little bit further, x plus, we get 1 eighth, the integral of 8x minus 4 over 4x squared minus 4x plus 3 dx. And then the last part, because we took this negative 4 already, so we have a negative 4. But if you times negative 4 by 1 eighth, that's just gonna be a negative 1 half. So we get negative 1 half, the integral of dx over this denominator 4x squared minus 4x plus 3. Like so. It kind of a little got cramped right there, but that's the part we need to deal with because this second integral, we know exactly what to do with it now, you're going to get just 1 eighth times the natural log of the absolute value of 4x squared minus 4x plus 3. Turns out we don't need absolute value since the thing is always positive, like so. And so what do we do with this last part? We have this negative one half, the integral, and now I won't squish it this time 4x squared minus 4x plus 3. So we saw in the previous video that you're going to have to proceed to do some type of trigonometric substitution. In which case if you do that, you start off with having to complete the square because it's not quite ready to go yet. And so if you have this 4x squared minus 4x plus 3, to complete the square, we're going to separate the x's from the constants. So we get 4x squared minus 4x plus 3. Take out the four. So you get x squared minus x plus 3 right here. Taking half of the middle term, which is itself one half, you're going to square that to give you one fourth. So we have to subtract four times one fourth to compensate for that. That gives us four times x minus one half squared plus three minus one, like so. Then I should mention here that for itself is a perfect square. You could write this as two times x minus one half quantity squared plus two. This has the advantage that you could write this as 2x minus one squared plus two. So now it took a little bit of effort, but we now complete the square. If you wanted to do a tangent substitution, you would then set 2x minus one equal to the square root of tangent theta, and you could proceed to go from there. But I actually have an alternative that might be better for us. And it might just be to memorize a very useful integral that's hard to do. So if you take the integral of dx over x squared plus a, this is the type of integral where you would set x equals a tangent theta. So be aware of that connection right there. So back up. This thing is going to always turn out to be one over a tangent inverse of x over a plus a constant. And it might just be worth memorizing this right here because our integral right here looks just like this thing when you complete the square. And so using this formula, let me copy down what we had before. You have x plus one eighth the natural log of the polynomial for x squared minus four x plus three. And so then we're going to get a negative one half sitting in front of Well, in this situation, you're you're going to replace x with a two x, right? So you're going to get our tangent of two x minus one over the a value a values give me the square root of two, this thing right here. So you get a square root of two and the inside and you get a one over the square root of two on the outside, plus an arbitrary constant. So if you memorize this formula, it can save you a lot of headache if having to go through the trigonometric substitution. Because in this situation where you have to do a trig sub for partial fractions, it does turn out to have this form. And then just times everything by negative one half, let's do it one more time. x plus one eighth the natural log of four x squared minus four x plus three minus one over two root two, our tangent of two x minus one always square of two, plus an arbitrary constant. So as we're going forward, these partial fractions decompositions because we're insistent on not using complex numbers, my recommendation is to just simply memorize this formula here in yellow, write it down in a notebook and use it because it'll be it'll be the trig sub you have to do all the time and it'll be your saving grace in the situation like this. Do the best you can it turns out you could have also started you actually started doing your trig sub back at this moment right here. Right, we could have done a trig sub here instead of doing this u sub and then thing over here. This thing over here I meant to say, you could do that. Now, if you if you if you have excuse me, if you have this thing memorized, then it'll be faster to do it the way we just did it. But if you plan to just reconstruct this from trig subs, it would have faster star with the trigonometric substitution. These partial fraction calculations can get very difficult when a trig subs involved, but I recommend that you try to memorize this formula in the yellow to help you simplify your calculation along the way.