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Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefficients.

(a) $ \dfrac{t^6 + 1}{t^6 + t^3} $ (b) $ \dfrac{x^5 + 1}{(x^2 - x)(x^4 + 2x^2 + 1)} $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Campbell University

Harvey Mudd College

Baylor University

Boston College

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Write out the form of the …

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I-6 Write out the form of …

Let's go ahead and find the partial. Freshen the composition for dysfunctions in part A and B in party. We see that the new writer has degree that's equal to the denominator. They're both degree six in any time. The numerator has degree greater than equal to the denominator. We should go ahead and use long division polynomial division. So we'LL have a one up here t six plus t cubed and then we have minus t cubed and then bring down the one that's our remainder because he did the six does not going to t cute Silin party, eh? We can go ahead and rewrite this as the quotient which was one up from a peer. Then we have our remainder one minus t cubed, all divided by the original denominator. So now we could go ahead and use a partial fraction to composition on this fraction. But I would always go ahead and factor out definitely on the bottom. But it won't hurt to factor in the numerator because every now and then you may have cancelation. That will make the problem easier. So let's go in factor. This this is just one minus t T square plus T plus one. So this is just a difference of cubes, a cubed minus bean cube. There's a formula for that, a minus B that was the one dynasty and then a square, plus a B plus piece for and then on the denominator. We could go ahead and pull out a T. Q. And we looked over with T cubed plus one. But this is a sum of cubes t cubed plus one cube. So there's a sum of Q's formula that you might recall this time. It's a plus B and then a square minus a B plus these flare so we could go ahead and write this as thank you t plus one t squared minus T plus one. And we see that there will be no cancellation here if you compare these circle terms. But at least we've simplified the denominator. What we should do here is check if this quadratic factors or not, because if this factors we need to factor before me too partial, freshen the composition. So recall that if you have a polynomial second degree quadratic, this will not factor. It's B squared minus four. A. C is negative in our problem for this polynomial. We see that A and C are both won CB is minus one So plugging in a B and C into this and inequality we have one minus four in a less than zero. So this means that this contract is in factor. So let's go ahead and rewrite this. This was our previous expression and now this is all equal to one plus. And since there's no factory ization and there was no cancellation, you could go ahead and just leave It is one minus t cubed, if you like. Doesn't matter. We'LL replace it in a moment that on the bottom we have t cute is he plus one and then t squared minus t plus one. I'm running out of room here Let me go on to the next page Let me rewrite what I had Oh, we had a he cubed t plus one Is he squared minus t plus one? So that was after we factored the denominator. So now we look at the first term we see t Q There. This is what the book calls case too because we have a linear factor t it's being raised to the third power. So he is the linear factor there that's being cute. So much is for the T cubed we'LL have Ah, a liberty deal is he square seal Retty cubed And again this A, B and C is coming from the formula given for case too. Then the next term corresponding to t plus one. This is just a case One term non repeated of linear factor So we'll have another constant over t plus one and then finally this quadratic over here that did not factor This is what the book calls case three and so we have e x plus f by the formula given in case three and sorry, that should not be except there. And that should be a That should be a TV because that's our variable. So should be a tea there e t plus, Then we have our quadratic and so for party. That's our partial fraction to composition. And as the problem mentioned, it was a necessary to find the constants a all the way up so we can stop right there. So it's box this and let me go on to the next page here for part B. So we have exit the fifth plus one X squared minus X X for two X Square, plus one in the denominator as well. So looking at that first term X squared minus X in the denominator, we should try to factor that. So there we can pull our X and we'LL have X minus one then for this term over here. Although it's fourth degree, it's technically a quadratic and disguise we can rewrite. This is X squared square plus two x squared plus one. Or if you want, let's replace X squared with another variable that this quadratic just becomes w Square plus two w plus one, which we know is just w plus one all in parentheses square and then going back to X. We can run it like this. So we have a X square plus one square down there, and our numerator still just accept his plus one. We're almost done here. The last thing to do is to look at this quadratic that's being squared and see whether or not this thing factors. So Justus before will check the D squared minus four a C condition here be a zero and then a and C, you're both won. So this will just be minus four. That's less than zero. This means that the quadratic does not factor into tool in years. Okay, so let's go to the final answer. So taking this whole expression, the next line will be well, the first term is X, so we just go to case one here its linear, and it's non repeaters. So we have the ever X for case one, then here. Still, in case one, it's another linear. It's different in the previous one, and it's non repeated. Some B over X minus one and then for the last term X squared plus one squared. This is what the book will call case for because it's quadratic doesn't factor, and it's repeated. So we'LL have CIA closed over X squared plus one. And because this repeated, we'LL need an extra term here e X plus f over X squared plus one square, and there's a final answer

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