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

## (a) $\frac{1}{x^{2}+x^{4}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C x+D}{1+x^{2}}$(b) $1+\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-1}$

#### Topics

Integration Techniques

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

here. Let's go ahead and find the partial fraction to composition once we do so it's not necessary to find the values of the coefficient here. So here's our fraction. The first thing we should try to do is look at that denominator and see if the factors we always wanted found there as much as we can. Here it's important. And this the nominated It looks like we could take out an X squared and we're left over with one plus x squared and we'LL still see if we can factor. So looking at this first term here, X squared. This puts us in what the book calls case, too. We have a repeated linear factor in this case. It happens to be X. So there's X and it appears twice second power. And then we'LL have to also look at this for dramatic here. We'd like to know if that factors that this might factor into too linear polynomial sze. So given a polynomial a X square plus B X plus C. If you'd like to know whether you can factor this thing using real numbers, this does not matter if B squared minus four A. C is less than zero. And our problem here the A, which is the constant in front of the X Square, is just one be a zero. There's no ex term, so be has to be zero. And we see the constant term C is also one and therefore B squared minus four. A c equals negative for this's negative number. So x squared plus one does not back there. And so this is what the book calls case story. This is when you have a quadratic factor that doesn't factor into too linear polynomial sze or what some might call an irreducible quadratic. So now putting case to in cases three together. So looking at the first term, the one over X squared Listen, be color coordinated here we have in Blue Rex because X is the linear factor. And then since X goes to the second power we do another term and we go all the way up until we hit the highest power of X and it just happens to be two. So we'LL stop there for the one over X squared And now for the quadratic using history, we'LL have that quadratic back on the bottom on the denominator, But then on the numerator, instead of having a constancy, we need to have a linear polynomial appear. So this is case three. So we'LL put a CX plus de So this is it could be any polynomial that's linear for part me. The first thing we should do here I actually have to go to the next page for part B because the numerator is degree three denominators degree three, We should do polynomial division. So let's go to the next page and do long division here. So let's rewrite that polynomial or this case the rational function X cubed minus one Execute minus three X square plus two x Let's go ahead and divide this What? So x cued times one is execute so well guarded multiplied this all by one and then we subtract and we're left over with three X square minus two works minus one and we can no longer divide because this is a larger power of X. Then we have down here. So this tells us that the original fraction could be written as one, and then we're left over with the remainder, and then we have the original denominator and now we can consider the partial fractions because now the numerator has degree That's smaller than the denominator. So now we tried a factor. The denominator as much as we can. And we could do the partial fraction the composition. So in the denominator, the first thing we could do is pull out of next term so that we have a quadratic left over and this quadratic wolf after in the denominator we have X and then we have X minus two X minus one. And now this is what the book calls Case one. We have linear factors that are all this thing. So we'LL have one linear factor. They get a constant on top. So we'll have this thing Constance A, B and C for each one of these terms and that's your answer for part B.

JH

#### Topics

Integration Techniques

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp