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JH
Numerade Educator

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Problem 31 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{1}{x^3 - 1}\ dx $

Answer

$$\frac{1}{3} \ln |x-1|-\frac{1}{6} \ln \left(x^{2}+x+1\right)-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}(2 x+1)\right)+K$$

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Video Transcript

Let's evaluate the given into rule here. We should see if we can pass through the denominator and we can. This is a difference of two cubes so we can write this as one over X minus one and then we'LL have X squared plus X plus one. And then we should look at this contract IQ in the denominator to see if that factors we can look for the discriminating B squared minus four a. C. Here, b is the coefficient in front of the X. It's just the one, So that's one square minus four Time's a. The coefficient in front of the X Square is one, and C is also one, So we have a negative quantity. That means that this quadratic will not a factor. And so now the partial fraction to composition should look like a over X minus one. That's what the author calls case one non repeatedly near faster. And then here we have Case three because we have and irreducible quadratic factor. And now well, just go ahead and find a B and C. So it's multiply both sides of this equation here by the denominator on the bottom left, we have one equals a X squared plus X plus one and then B X plus C and then X minus one. And let's just go ahead and expand this ready inside X squared eggs. Hey B X, where fear minus B X minus e and then let's combine like terms. Let's fact throughout X squared. We see that there's a plus B was for throwing ex. That leaves us with a plus, he minus B and then our constant Herm left over a minus c. So this will give us three equations and three variables. So a plus B we see that there's no X Square term on the list. That means a plus B a zero. Similarly, there's no X term on the left. There's just the one there. So that means that a plus Eamon is B is also zero. And then the constant sir, um, on the raid must be one to match the one on the left. So now we have a three by three system that we could solve for a B and C, for example, the first equation. You could solve that for, eh? Then you could plug this a value into the other equation. here. The second one, it's C minus who be equal zero. So C equals to be. And then from the last equation, Yeah, A was one plus e, but C is equal to to be. But then to be because of this over here equals negative too, eh? So we can solve this for a we get three equals one So equals wonder then that gives us b equals negative. A third from the circle equation over here on the left and then see is two times b. So we have C equals negative two thirds. So those are our Constance for A B and C will, which is go ahead and plug those in up here in the upper rate. And then we're ready to integrate. Let's go to the next page. So in a rule can be written plugging in our values for a B and C, of course. So that's the be in the sea. And then we had the quadratic. Now we could split this into two, actually, three rules here, So let's go out and distribute this integral sign. We could Claude the one third one over X minus one, and then for the next term. Let's actually just leave that. Let's just chlo aa minus Wonder here and I'll just have a X plus two and then for this denominator here let's go ahead and complete the square. So that will be X square plus X. And then there's a one in the front Take half of that and then square and then make up for this by subtracting the one force So we get X plus a half swear plus three over for So that'LL be our denominator here After we've completed the square let me write that as room three over too squares you already that way So that in case we have to use tricks of it'LL be easier to see what that Ames. So for the first integral, this is just wonder Natural log Absolute value X minus one. Now this minus one is bothering you here in the bottom. Feel free to do it use up and then for the other interval here you may go ahead and try tricks up right now, what might also work here is to rewrite the numerator. So let me write this as plus a half. So I'm doing here is writing too as a half plus three half And the reason are pulling on the half is because I see X plus half in the denominator So let me just leave the one half here. I'Ll deal with the three house in a second. So that deals with the one half here and now for the three house I'LL just come down and write that on the side So I still have this minus one third That was outside the integral and then just three halves up here again. This wasn't the only way to solve this problem. You could've used tricks up right away. We'Ll still use tricks up in a moment, just simplifying the problem and DX Now we've already evaluated the first integral here we have our natural life. Now we have two more one in the red, one in the green. The reason for pulling out the X plus one half was the that we can use a U substitution for the red one. So for this integral and red, we should take you to just be in this case, just take you to be the whole denominator. Three of the four and then do you this ship over to do you over to is just X plus a half d X. And that's exactly what's in the numerator. So we can go ahead and write this integral in Red is negative one third and then we have a do over too. So that should be there, not forget the one half and then all we have left over is one over you. So this is just negative. One over six Ellen absolute value. You negative one over six Natural log and then just replace you with this term over here. Explicit have squared plus twenty fourth and this can also be written is just we read down here Just a nicer looking expression here X squared plus X plus one just simplifying, that's where. So we have two of the three animals we have one more and greed for this one. The use of just won't work out is nicely here because it's still will require the troops up. So in this case, let's just go straight to the tricks up. Let's go to the next page to do that. So we had for the green and rolled the last one, and in this case, let's go out and cancel those threes, Actually. Sorry about that. So we're one half and then we have DX explosive Have squared plus room three over two square. There we go for a transom X plus a half equals room three over to tan theta, then DX it's fruit three over to C can't square. So let's go back and write are integral Oh, one half so D x three over too. See, kings were data and then in the denominator we have X plus a half. So that this term over here and we'LL Square So we have three of her four tents where plus three over four. So this denominator pullout of three over four you get tan Square plus one and we know Tan Square plus wanna see can't square so we could cancel those terms. So then we could go ahead and simplify these fractions and we just have after canceling, we have another two here and then that four will go on the numerator so the force will cancel negative room three And then now we have a three in the bottom as well. The sea cans canceled, so we're just left with the data and roll that just a little, And then at this point, use your tricks. Upsets offer data so theta equals arc Tan explosive half over. Room three over to. And let's go ahead and simplify these fractions. It's getting a bit messy here so I can write. This is ten in verse two X plus one over Route three. It just got a common denominator of two on the numerator and denominator and then cancelled its use. So this is the final inner rule, and we're back in terms of X now, So the last step will be so she's go ahead. Not This won't be our final answer. That was the last, the third in a girl Now the last f. Let's just go ahead and add those three and the girls together. Let's go to the next page to write that out. We had natural log X minus one that was divided by three. Then we had minus one sixth natural log X squared plus B X plus one, and then finally, for the last dinner girl, using the truths of new three over three. It's an inverse two x plus one over Moon three. And don't forget our constancy of integration, and that's your final answer.