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Use partial fraction decomposition to evaluate the integral: 18 dt = t2 13t + 22 PreviewBe sure your variables match those in the question....

Question

Use partial fraction decomposition to evaluate the integral: 18 dt = t2 13t + 22 PreviewBe sure your variables match those in the question.

Use partial fraction decomposition to evaluate the integral: 18 dt = t2 13t + 22 Preview Be sure your variables match those in the question.



Answers

Use partial fractions to evaluate the given integral. $$ \int \frac{4 t^{2}+3 t-1}{t^{3}-t^{2}} d t $$

Clinton. We have to integrate this using the partial fractions. So first we need toe factories the denominator completely. So let's, uh, split the denominator. Factors the denominator you by splitting the middle term scan. Britain as X squared plus nine x plus two weeks plus 18. This can be written as expressed two times. DX over. Let's take X out from here and to walk from here so we're left with X Times X Plus nine. They left for two Times X plus nine Integra on their did from the denominator. Again, we can take Express nine out, so we are left Foot X plus nine times X plus two. Fortunately, this factor gets cancer and we're just left for D X over X Plus nine, which is a bigger law which will be up Ln off absolute value of X plus nine plus the constant of ridiculous. See

We're going to start with a substitution. So we will let X be equal to eat to the T. Then dx is going to be E. To the T. Mhm. And then we're actually going to isolate GT. So D. T. Is going to be one over each of the tea. But either the T. S. X. So that's really just one over X. D. X. Okay so let's go ahead and substitute that in. So we get this is the integral of one over. So eat the two T. That's X squared plus one And then D. T. is one over X. Yes thanks. So if we actually multiply this through we get that this is equal to the integral of one over X. Times X squared plus one dx. And that is something that we can decompose. Okay so let's do that here at the bottom. So this is one over X times extra plus one. We actually already have the two factors because one is X and the other is X squared plus one. So our numerator here with the A. And the explicit Okay so now let's multiply this through and compared enumerators. So we have a times X squared plus one. That will be a X squared plus A plus X. Times X plus C. So let's be X squared plus C. X. Okay now we can compare like terms the terms with X squared would be A plus B. And since we don't have any X squared term on the left hand side. And that must be equal to zero. And the term with just X. That C. And we don't have any terms with X on the left hand side. So again he is just zero and then finally the last one is A. Which is a constant term and that must equal to one. Okay so which means if A. Is equal to one, B must be equal to negative one. Okay so that means we can now rewrite this as two separate integral. So it'll be the integral of A over X and A is just one. So this is one over X. The X. And then plus or since B is negative one who can just write this as minus the integral. Uh X over X squared plus one. Okay so this is equal to L. N. Of X. And this one here we can do a substitution because if we let you be equal to X squared plus one we get the U. Is equal to two X. Dx. So therefore this one is equal to the integral of So we have a one half in the front and this is just one over you. D you Okay so we can go ahead and do that. So this is Ln of x minus one half of Elena View and we can just put an expert plus one back into you because you can substitute it back and now the last part is willing to substitute either the T back into X. Okay so therefore our final answer is going to be Ln of E to the t minus one half. Ellen Sorry. Earlier, I forgot to see. So let's put that in and this is our line of E to the to T. Plus one. Okay, you can simplify this just a little bit more because our line of either the T. Well that's just T. So our final answer is going to be t minus half of Ln of E to the two T plus one plus C.

For this problem we are to evaluate the given integral. Using partial fractions, we begin by finding the partial fraction decomposition for t minus one over T to the fourth power plus 60 cube plus 90 squared Now, -1 over T to the fourth power plus 60 Q plus 90 square Can be written into T -1 over the square of T times the square of t plus three. Now, since we have repeated then you're factors then are partial fraction. The composition will have partial fractions with denominators that our powers of T and T plus three. So we have a denominator of T, T squared T plus three and the square of T plus three. So since all of these denominators are linear, then the corresponding enumerators are constants. That's A, B, C and D. As you want to find the values of the coefficients, A, B, C and D. By multiplying this equation by the LCD, t squared times the square of t plus three. So from here we have T -1 equal to eight times tee times the square of t plus three plus b, times the square of t plus three plus C, times the square of t. Time's D-plus three plus D times the square ft. Actually want to fix values for T by setting The denominators of the partial fractions two, Zeros and here we want to let T equal to zero and we will use this x value two, evaluate our equation and and here we want to have 0 -1 equal to zero for the first time since we have a factor of tea which is equal to zero. Plus we have B times zero plus three to the second power plus the third term will be zero. Since you have a factor of teachers equal to zero also for the last term. So simplifying this, we have -1 equal to nine. B. Which means that Base equal to negative 1/9. As you want to set T plus three equals zero. So if T plus three equals zero this means that equals negative three and our equation becomes -3 -1. That's equal to for the first time it becomes zero since we have T plus three X factor Plus the second term will also become zero As well as the third term. So we have plus zero and we have plus D times The Square of -3. And this gives us the equal to Negative for over nine. Since the remaining denominators are just powers of T. And T plus three. Then we will randomly pick values for tea. Let's say T is equal to -1 is then Our equation becomes -1 -1. This is equal to eight times negative one times the square of negative one plus three. Plus Beaches negative 1/9 times negative one plus three squared plus c times negative one squared has negative one plus three plus. Do you just negative 4/9 times the square of negative one. And simplifying this we have -2 equal to negative for a -4/9 plus to see -4 overnight. This gives us four a minus. To see that's equal to 10/9. And if let's say it is equal to one this gives us one minus one. This is equal to eight times 1 times one plus three squared plus B. Which is negative 1/9 times one plus three squared plus C times one square times one plus three. Plus we have details just negative for over nine Times The Square of one. Simplifying this we have zero equal to 16 a minus 16/9 plus four C -4/9. That simplifies to 16 8 Plus four C. This is equal to positive 20/9, dividing the Equation by two. We have 80 plus. To see This is equal to 10/9. Let's call the equation for a minus two C. Go stand over nine our equation one and let's call 88 plus two C equals stand over nine. Our equation too. Now if we add a regulation one integration to we have four a minus to C equal stand over nine. This added by the equation 88 plus two C equals stand overnight. This gives us 12 a equal to 20/9. And if we divide both sides by 12 we get a that's equal to 5/27. Now if AS 5/27 then substituting this either two equation, water equation to we have four times 5/27 minus two C. This is equal to 10/9. This gives us 20 over 27 minus to C. Equal to 10/9. That means to see is equal to negative 10/27. Which means that C. Is equal to -5/27. Now that we have values for A. B, C. And D. Then the partial fraction decomposition for T -1 over Teacher. The 4th power plus sixties To the 3rd power plus 90 squared. This is equal to 5/27 over T. Plus we have -1/9 over T squared Plus negative 5/27 Over 2-plus 3 Plus negative for over nine over the square of T plus three. Now simplifying this, we have 5/27. This times one over to you -1/9 times T race too negative too -5/27 times one over T plus three -4/9 times D plus three. Race too negative too. Now integrating both sides with respect to T. We have the integral of t minus one over T. To the fourth power plus 60 Q plus 90 square D. T. This is equal to 5/27 Times The integral of one over T -1/9 integral of T. Race to negative two D. T minus 5/27 times integral of one over T plus three D. T. And then we have -4/9 times integral of T plus three. Race too negative too. DT simplifying this we have 5/27 times L. And absolute value of t -1/9 times. He raised the negative one over negative one -5/27 M. L. An absolute value of T plus three -4/9 times T plus three. Race two negative one Over -1 and then plus C. So from here we have 5/27 times. Ln absolute value of T over T plus three minus a negative value becomes plus 1/90. Plus we have 4/9 times T plus three and then plus C.

Hello. This video will show you how to use partial fractions to integrate. With this complex fraction, we can see that the denominator is cubic. Therefore, when we factor this, we're going to see that we have three terms. In other words, we're gonna have three partial fractions. So first of all, factoring at one layer at a time, I'm going to take about that J c f of tea. But that'll leave T square T minus two. Doctoring this one more layer, I'll have tea team plus two and T minus one thistles. My fraction that I'm going to split into three parts. So I'm going to be looking for A B and C A will have the denominator t b will have the t plus to you and see Will have the T minus one. Aren't multiplying both sides by that common denominator of T T plus two. You and T minus one. That'll leave me nothing but one on this left side. A will be multiplied by the T plus two on the T minus one B will be multiplied by t and to you minus one. He will be multiplied by the tea and the T plus two. All right, so lots of distribution here. I'll have a T Square. I mean, be witness to a with the baby. I'll have be he squared minus beauty And lastly see t v squared Plus two de t. All right. Collecting all my terms together, I see three of them with the t squared term. All right, so that'll be a plus B plus c t squared. Next, I'll get my teas together, So that will be hey minus b plus to see, but he's down. Lastly, I have a negative to A for my constance, which, it turns out, that's all I have. Um, in my numerator is the one. So that's the good news is that we're gonna have to equal the to a or A is a negative one. Half the sea is gonna take a little bit of algebraic manipulation. So let's see if I've got a plus B plus c, that's gonna need to be zero, cause there is no t squared term. But I already know that a is a negative 1/2 so negative 1/2 plus the B plus the sea needs to be zero in the same way. Looking at my key terms, I have a minus B plus to see again A is a negative 1/2 minus B plus to see again. This needs to equal zero. So let's go ahead and add these two equations together to cancel out my bees. If I Adam together, that will give me negative one three C's equals zero Sell their it iss c is gonna be 1/3. All right, almost there. I need to find b. So using one of those equations, Maybe that 1st 1 negative 100. You know, I'm still looking for this. Will give me B is gonna need to be a 16 All right, so there's my a, B and C. What this means is that back to my original integral of one over t cute peace squared minus two t that we can instead do the integral off a is my negative 1/2 over t ah, plus B is the 1/6 it's with the T plus two. And lastly, see is the 1/3 which is with T minus one. So these three smaller fractions are gonna be much easier to integrate than that original. That'll give me your negative. One natural log 16 naturally have three close to use and 1/3 patients use here. That's just but we can finish this up. Um, I'm out the natural log. Sorry. Natural log of T minus one plus c No city. They're fitted in ISS, my girl. Thanks for watching.


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