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JH
Numerade Educator

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Problem 69 Hard Difficulty

(a) Use a computer algebra system to find the partial fraction decomposition of the function
$$ f(x) = \frac{4x^3 - 27x^2 + 5x - 32}{30x^5 - 13x^4 + 50x^3 - 286x^2 - 299x - 70} $$
(b) Use part (a) to find $ \int f(x) dx $ (by hand) and compare with the result of using the $ CAS $ to integrate $ f $ directly. Comment on any discrepancy.

Answer

(a) $\frac{24,110 / 4879}{5 x+2}-\frac{668 / 323}{2 x+1}-\frac{9438 / 80,155}{3 x-7}+\frac{(22,098 x+48,935) / 260,015}{x^{2}+x+5}$
(b) $\frac{4 \mathrm{s} 22}{4879} \ln |5 x+2|-\frac{34}{323} \ln |2 x+1|-\frac{3146}{80,155} \ln |3 x-7|+\frac{11,049}{200,015} \ln \left(x^{2}+x+5\right)$
$\quad+\frac{75,772}{260,015 \sqrt{19}} \tan ^{-1}\left[\frac{1}{\sqrt{19}}(2 x+1)\right]+C$

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Video Transcript

suffer party. Let's use a computer algebra system to find the partial fraction to composition. So answer this fraction into a computer algebra system. So here's Wolfram Alpha with input, the freshen. And then if we scroll down, we see a partial fraction to composition that's highlighted here. So let's go ahead and write all this on the white board. So for her, eh, we can rewrite f is the following. So we do have a lot of writing here with the large numbers. So this is using the computer algebra system. Tau. Give us the partial fraction to composition. So there's one term so that we have six, six a three, two, three, two worthless one also minus nine, four, three, eight, eight zero one, five, five And then we have one more term after this. So this is just for party here in one more term, twenty four one zero one one zero for eight, seven nine. Finance clothes, too. Now, the next part will be to actually God and integrate this. So the first integral is going to be the most difficult because of the quadratic. So what? To go ahead and complete this wear here you can write this as experts a have square, the whole thing and then radical nineteen Over too square. So we just completed the square there. Let's spend a few moments just on this first Integral. The other three hundred girls will just involve natural log And you could do them all by U substitution. So here let me take this first freshen and then just go ahead and write this. So what I'LL do here is I see that I haven't x plus one half on the bottom. So I'll rewrite this x It's X plus one half But by doing so, I have to subtract. So we have to to let me let me take a step back here That's two X plus one half and then we still have the plus forty eight thousand. But since I just added in this extra half, I have to make up for by subtracting one half of this number and then now I'm you just split this into two fractions. I still have that two hundred sixty thousand out there And now, after I completed the square, I have this and then route one nine over two square that as well So that's the first in immoral and the same denominator will show up in the infraction. Little sloppy here. And then we could even put a d s up here because we're integrating now was break this into two parts. So here we should take you to just be explosive. Have take it to be the denominator there without this number in the front. And then you over too. Expose the have d x so that in a rule, becomes it's then from our use of we have one half one over you We'LL see you there so this becomes it's a four up there in the front for nine Then we're natural log of you and then we could replace you with and swear was X plus five was put that in absolute value. So this right here is just Ellen ofyou and I just replace you with definition here on the simple side. So that's the first in a brawl. So remember, on the previous page we have foreign rule severe with we're just dealing with the most difficult undergo here that we circled in red. We split in the institute hearts. We just found one of the parts. We're one more parts of fine. And then we'LL simplify the other three simple intervals. Position of rule. Let's do trace. Um so go ahead and evaluate this. Using this truths up in the same manner that I did over here using the use home in this in a roll he comes seven five, seven seven two, two hundred sixty thousand We have that nineteen on the bottom and then our tan for this is from the tricks up two X plus one and then on the bottom. Radical nineteen. So the first most complicated in a girl that we had we just broke that into two parts. Now we also so we'll just add those together at the very end And then we had the second and rule six, six eight over three to three two x plus one. You're free to do a use up here. Then the next interval. The third one, who's a eighteen to be X minus seven gs. So go ahead and use up here as well. You forced three x minus seven You get three one four six He's zero one one one five five. Natural log three x minus seven and then for the final in a rule. So here's our last Enbrel. Then we find we don't get him at everything all of the end. So go ahead and use of here equals five plus two. Simplify this integral to the natural world. By that plus two, the absolute value and other final answer was It's just the interval, the sum of the two hundred girls that we had in the previous age, then minuses in general, minus the green in a brawl. And then plus this last read and mumble. So let me go to the next page to write that. So our final answer it's And then after we write this down will go head backto Wolfram to the computer algebra system, and we'LL see if they're any differences in how our final answers there. My guess is that they just won't have absolute value inside the natural, and we'LL check it momentarily. Let's keep writing this answer down, adding, It's attracting all these in a girls together, so that's the sum that's the most complicated. And a girl. That's what we had used to usurp amended tricks up. Then the next, in a brawl them we subtracted there following in a role as well. So it's a one five five ln three x minus seven and then the final interval, we added that was for a two two. So this is our last expression here. Natural log five x plus two. And finally we could go ahead and add in that constant of integration. See? So this whole page, this whole y borders are final answer and let's go ahead and compared to Wolfram. So no, the constants not know. Also, we have absolute value here for all of our loves. If you go back to Wolfram, our computer algebra system, let's scroll down and it gives us the indefinite in the rule of our friendship. It's just write it in a different order, but it is the same fraction. And if you look inside, the longs hereby log, They mean natural other of them, as they mentioned on the side. But they do not have absolute values, but otherwise our answers are the same. So that's your final answer.