Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Partial fractions taking a fraction and breaking it apart into more simple pieces that can then be integrated. So an example, if I have five X - 7 / X ^2 - 3 X +2, and I want to break that up into two fractions that add together to give me that. So the first thing we're always going to do is we're going to factor the denominator as much as possible. So in this case, we have X - 2 and X - 1. So we want to find A&B where a / X - 2 + b / X - 1 is going to give us the original fraction we started with. So if we think about getting a common denominator, we'd have a * X - 1 + b * X - 2 all over the common denominator. So then we could see that the numerators would have to be equal. So five X - 7 would have to equal a * X - 1 + b * X - 2. Now one way to solve for A&B quickly is if you can see that one of the terms, the a or the B portion will go to zero with a value for X. We could substitute that in. So if I let X equal 1, I get 5 * 1 - 7. Equaling a 1 - 1 is going to give us 0, so 0 * a plus B 1 - 2. If we just simplify that up, we can see B equal 2. Then we would do the same thing by letting the X equal 2 to make that second term go away. So if we stick to and for X, we get a equal 3. So we would have the fraction 3 / X - 2 + 2 / X - 1. Now the problem is that that method does not always work. A different method is to set the two numerators equal and distribute things out and then realize that all of the XS on the left side have to equal all the XS on the right. So the coefficients on the X the five would have to equal the A+B. All the constants on the left would have to equal all the constants on the right, or -7 would have to equal negative a -, 2 B. Now when we solve this, we could add these two equations directly, seeing the A's will cancel -2 equal negative BB equal 2 if I know BI can substitute it back into either of the two equations to get A. That method actually will always work. The first method will work sometimes. The second method I just showed works all the time. So now if we wanted to take the integral of that original partial fraction, so we have the integral of five X - 7 / X - 2 * X - 1 DX. We didn't know how to integrate that, but now we can rewrite it as two separate fractions. So the integral of 3 / X - 2 + 2 / X - 1 DX. And we use our laws of integration and we know that that's just going to be 3, the natural log of the absolute value of X -, 2 +2, the natural log of the absolute value of X -, 1 + C So then we could take the three into the exponent and the two in the exponent, and we can multiply those two natural logs together. So we'd get the natural log of X - 2 ^3 * X - 1 ^2 + C the absolute value of all of that. If we look if the bottom is a linear term or a learning linear term to a power, then a constant goes on top. If the bottom is a non factorable polynomial then the top is 1° less than the bottom. So if I have X ^2 + 1, I need AX plus B on top 1° less. If I had X ^3 -, 2 X squared +4, that's not factorable. So I'd have AX squared plus BX plus C on top. If we look at two X + 2 / X ^2 - 2 X plus one, when we factor this X ^2 -, 2 X plus one, that's really X - 1 quantity squared. So what we're going to do is we're going to rewrite it as a / X -, 1 and b / X -, 1 ^2. The reason we're doing this is it's a linear term. Once we look at the factorization, it's a linear term. So it's constant on top, but it's a linear term squared. So we need every power up until the power that was squared. If it had been a cubed, we'd have the first degree, second degree, third degree. If it had been the 5th degree, we'd have first, second, third, 4th, and 5th. So getting a common denominator, I can see a is going to be multiplied by X -, 1. The B is already over the common denominator and that had to equal the original. So the two X + 2. If we let X equal 1 in this case, the a part would cancel so we'd get B equaling 4, so a different method. Still, once we got our B, we didn't have anything to let X equal to get rid of the B to find the a. So we could take the derivative of each side. We know that the derivative of each side with respect to X has got to be equal. So if I have AX minus A+B for this side here the derivative of this is going to equal the derivative of the right side. So the derivative of this is a constants negative. A&B has to equal 2, so we could have 2 / X - 1 + 4 / X - 1 ^2. When we go to integrate this, we would have the integral in terms of DX and the integral of this whole thing in terms of DX. So we get 2 natural log X -, 1 -, 4 over the quantity X - 1 + C because remember we could think of that for X - 1 to the -2. And if we add 1 to the exponent and divide by the new exponent, that's how we got this term here. It's a bubble thought. So another way back here we could have done the coefficients that does always work is if we have AX minus a + b equal two X + 2, make the coefficients on the X term equal the coefficients on the X term and then make the constants equal the constants. So then we can see if a is 2, we put a -2 we get B is 4. Looking at this next example, the degree on top is more than the degree. Actually it's equal. If it's more or equal, we have to do long division. So T to the 4 + 9 T squared divides into T to the 4 + 9. T to the four goes into T to the four one time. So 1 * t to the 4 + 9 T squared is going to give us T to the 4 + 9 T squared. We usually change our signs so that we can add or we could think of it just a subtraction. So the T to the four is canceled, we get -9 T squared and we bring down that 9. Well, T to the fourth times. Nothing gives me negative T9 t ^2. So that's going to be our remainder. So we're going to get this integral is really the same thing as the integral of 1 + -9 T squared plus 9 / t to the fourth plus 9 T squared DT. So now we're going to take and we're going to split up this t ^2, t ^2 + 9. I'm going to factor that denominator. This just factors pull out AT squared and we get t ^2 + 9, which cannot factor any further. So we're going to have t ^2, t ^2 + 9 in the denominator. So the one just keeps going along. We're going to get a / t + b / t ^2 because the t ^2 is really a linear term that's squared, so a constant over the TA constant over the t ^2. And then because this is t ^2 + 9 and it can't factor, we need a linear on top of there. So CT plus D all over t ^2 + 9. I'm just going to ignore the ones for a minute until I actually get into the integration. So I'm going to have this negative nine t ^2 + 9 equaling getting a common denominator on the right side AT t ^2 + 9 plus BT squared plus 9 + t ^2 times CT plus D. If I multiply that all out, I get negative nine t ^2 + 9 equaling AT cubed plus 9AT plus BT squared plus 9B plus CT cubed plus DT squared. Now the cube coefficients on the right have to equal the cubed coefficients on the left. There aren't any cubes on the right or on the left, so that's 0. The cubes on the right are A and C so this a + C because those are both on the cube terms have to equal the cube term on the left and there wasn't any, so that's a zero. Then the squared terms so -9 would have to equal the square terms on the other side AB plus AD so -9 equal b + D the non squared terms. Hence the linear term on the left there isn't 1. So that zero has got to equal the linear terms 9A on the right and then the constants on the left have to equal the constants on the right. So 9 has to equal 9B. Now one of the ways that I help myself as I count how many terms here 123456 and I need to make sure I had six of them here. That just makes sure that I haven't lost any. If I have two on the left, I needed to have a 2 total other than the zeros on the left. So now we can see that a had to equal 0B had to equal 1. If a is 0, then by this top 1C had to be 0, and if B is 1 then D would have to be -10. So we can rewrite that original integral now as the integral of one plus the a was 0, so we're going to ignore it. The B was 1 / t ^2 minus the C was 0. The D was 10, negative 10. So plus a -, 10 / t ^2 + 9 DT. So we can easily integrate one that's just T and we can easily integrate 1 / t ^2 that's -1 / t. Now this 10 / t ^2 + 9, we could think about using our trig substitution. That's just going to give us a triangle. It's t ^2 + 9. So the T is going to be on the opposite side. The three is going to be on the adjacent side so that we get tangent Theta equaling t / 3 or three Tangent Theta is T SO3 secant squared Theta D Theta is DT. So if we just look at this last piece of the integration, we'd have -10 / t ^2 which would be 9 tangent squared Theta +9. If we factor out a nine, we'd get tangent squared Theta plus one, which is really just secant squared Theta. Instead of the DT, we'd put in three secant squared Theta D Theta. So the secant squares are going to cancel the three and the 9 are going to reduce, and we're going to end up with -10 thirds D Theta or -10 thirds Theta when we integrate plus C plus C. And then to figure out our Theta coming up here, if tangent Theta is t / 3, then Theta is tangent inverse of t / 3. So we get -10 thirds tangent inverse t / 3 + C So putting that into our final answer would get the t - 1 / t - 10 thirds tangent inverse t / 3 + C We can also do these problems as initial value problems. And so we're going to get all of the TS on one side and all of the XS on the other. And then we're going to integrate each side. So we're going to get the integral of DX equaling the integral of 1 / t ^3 or t ^2 - 3 T +2, which factors into t - 1 T -2 DT. So when I integrate each side, I get X equaling. We don't know how to do this. So we're going to use our partial fraction idea. We're going to get A / t - 1 + b / t - 2 to equal the one over t - 1 T -2. So getting a common denominator a * t - 2 + b * t - 1 equal 1. So the T's on one side have to equal the T's on the other. So A+B would equal zero -2, A-B would equal 1. When we solve this, if I add them exactly the way they are, the B's will cancel. We get negative A equal 1 or a is -1, and if a is -1, substituting it back into either of the equations, we can see B as one. So this is going to equal X equaling the integral of -1 / t - 1 DT plus the integral of 1 / t - 2 DT. So when we find the integrals of that, we get X equaling the negative natural log t -, 1 plus the natural log of t - 2 + C Now we're going to put in the initial value. So when T is 3, the X was 0. So 0 here equal negative natural log 3 - 1 is 2 plus the natural log 3 - 2 is 1 + C natural log of one is just 0. So we get C equal the natural log of two. So our final answer would be X equal the negative natural log t -, 1 plus natural log of t - 2 plus natural log of two. If we wanted to, we could rewrite that and we could have it be the natural log of t - 2 / t - 1. And we could actually, because this is a plus, we could make it X equal the natural log of two t - 2 / t - 1. Thank you and have a wonderful day.