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JH

# 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) $\dfrac{x^6}{x^2 - 4}$ (b) $\dfrac{x^4}{(x^2 - x + 1)(x^2 + 2)^2}$

## (a) $$x^{4}+4 x^{2}+16+\frac{A}{x-2}+\frac{B}{x+2}$$(b) $$\frac{A x+B}{x^{2}-x+1}+\frac{C x+D}{x^{2}+2}+\frac{E x+F}{\left(x^{2}+2\right)^{2}}$$

#### Topics

Integration Techniques

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##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University  ##### Kristen K.

University of Michigan - Ann Arbor ##### Michael J.

Idaho State University

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### Video Transcript

Let's go ahead and find the parcel. Fraction the composition for these fractions given in party and be so let's start with party except the six Santa X squared minus four on bottom. The first thing that always check for this type of problem which by this problem, I mean polynomial divided by polynomial. We always want to know if the numerator has degree greater than or equal to the denominator. If it does, we should use long division here. Numerator has degree six Denominator has degree too. So we go for the long division. So it's going to the right and do this. We have exit the fourth over here and then I'LL give us extra sixth minus for us to the fourth power Go ahead and subtract And then up here we get a four x square minus sixteen x where no, subtract that we get sixteen x square And then that'LL give us a plus sixteen up here So that will give us sixteen X square minus sixty four so suppressed that ever left over with sixty four. So that's our manor suffer party. We can rewrite this long division give us quotient of X to the fourth for X Square plus sixteen. Then we have our remainder. And then we have the original denominator explored minus four. And now we just go ahead and use partial fractions on this on ly remaining fraction over here. And now we know no more long division will be needed because the nominator has been a degree. So we should take this denominator and go ahead and write that as X minus two always want to check if we can factor these and then we have two distinct linear factors. So this will be what the book caused. Case one. So we'LL use the formula for case one to rewrite the fraction and this will become. We have a over X minus two and then another constant be over X plus two and that will be an answer for party. Let's go on to the next page for party party. So we re write that down for part B. We have excellent for O r. And then we have X squared minus X plus one x squared plus two squared. So we will not need long division here because the numerator has degree for where's denominator It will have degree six. So this will be degree for when you square it and then adding two more exes From this leading term, we'LL give you extra the six So no long division. So now we're ready to go straight to the partial fraction the composition. But since these air quadratic ce and the practices, we have to shut off the factor So let's look at X squared plus two. So if polynomial looks like this x x squared plus B X plus c, we need to look at B squared minus four a. C. If this is less than zero, then there's not a factor. The contract is not a factor. So for this polynomial X squared plus two, we have a close one. B zero c equals two and we see that b squared minus four a. C we'LL just be zero minus four times one times two so minus a which is less than zero. So this was not a factor, So that's a quadratic that doesn't factor. And then we could also look at this other quadratic over here. So this one, we'd raise that we have X squared minus X plus one. So for this one, we see a equals one B is minus one. She is one. So these where minus four A. C. We have one minus four times one times one. So we have minus three, which is also negative. So that means that this other quadratic on the bottom left also just impacted. So this is will be what the book calls. He's three. We have this think quadratic factors that irreducible that cannot be factored into smaller parts. So this weaken right is X plus me over X squared minus X plus one. So that's for Case three. And in this term, actually we'LL be using case for for the second term because this one's to the second power. So for this one, because it's a quadratic on the inside we'LL have a CX plus de over X Square plus two a sense It's a repeated factor We'LL have to do this once war but this time on the X squared Plus for two we'LL have to square the very last one and that will be our partial fraction to composition for Barbie

JH

#### Topics

Integration Techniques

##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University  ##### Kristen K.

University of Michigan - Ann Arbor ##### Michael J.

Idaho State University

Lectures

Join Bootcamp