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Evaluate the integral.

$ \displaystyle \int_2^3 \frac{x (3 - 5x)}{(3x - 1)(x - 1)^2}\ dx $

$$

\int \frac{-5 x^{2}+3 x}{(3 x-1)(x-1)^{2}}=-\frac{1}{3} \ln 5-\ln 2-\frac{1}{2}

$$

Integration Techniques

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let's evaluate the definite integral from two to three. So we see that the denominator is already factored for us. And the partial fraction to composition should look something like a over three x minus one. Because this by itself it's just a nun. Repeated linear factor. So that's case one. And then for X minus one square. This is a repeated linear factor, so we have a constant over X plus one and then another constant over X minus one square. Let's go ahead and multiply both sides of this equation here. Bye, Just denominator and the bottom left. So the left hand side, we're just left over with X three minus five x, but on the right, a X minus once Where B X minus one three X minus one and then see three x minus one scored and simplified it. Both sides here on the left. Negative five X squared plus three x on the right. Disquieting. Multiply everything out and that for B well three X Square, minus floor X plus one and then for see There's nothing to do there. So on the ray inside, let's go on factor that square. When we do that we have a plus three b, then with polo necks. We have negative too, eh? Minus four B. But then plus three c And finally, for the constant term. What a plus me Minus e. And this is where we'LL obtain or three by three system. On the left hand side, in front of the X squared, we see a minus five. On the right hand side, the corresponding term is a plus three b. So those are equal. And on the left hand side, in front of the ex, we see a three on the right hand side minus two eight minus four b plus three c. So those are equal and finally, the constant term under left zero On the right hand side, it's a plus B minus C. So those off also have to be equal. So this will give us a three by three system. Let's write this on the next page and then go ahead and solve it. They're a plus three b equals minus five, minus two a minus for B plus, Teresi equals three and then a plus B minus C equals zero. So many ways to go ahead and solve the system one way sausages. Take this equation over here. Circled and multiply both sides by three. Still, we have three a plus three B minus three C equals zero. And then we could just go ahead and cancel the seas by adding these two equations. So when we do so we're Let's over with a minus. B the seas cancel and then three plus zero is zero. Excuse me, history. And then we can Now we have a two by two system in Andy. So bring this equation back down here. A plus three b equals minus five. So it's going and subtract these equations. So now these cancel we have minus four b and then we'll have a three and then those double negatives that becomes a plus five. And that's a that gives us be divide both sides by minus four and then we can go ahead and software, eh? It was three plus b coming from this equation here. So then we have three plus minus two. That's just equal to one. And then to solve for C, we could rely on this equation up here. Top rate, so plugging in a we have won plus B, which is minus two and then minus C equals zero. So then we have negative one equal see? So we have all three values for a B and C. Now let's plug these into the partial fraction to composition that we had on page one. So let me go on to the next page to do just that and are integral now becomes integral to two. Three we have are partial fraction to composition. Now we plug in our values for a B and C, so plugging in a that becomes a won over three X minus one Now be was negative too. So we have a negative to let's just pull the minus side and then X minus one, and then we have minus one over X minus one square. So that was our because their value for sea was minus one. Now we're ready to generate for the first term my help you here to do a u sub u equals three x minus one. Here, you Khun, do you equals X minus one. And here you could also do X minus one. So you may not need to use up here, but after integrating, we should have won over three Natural log. Absolute value of three X minus one minus two. Natural log. Absolute value X minus one and then plus one over X minus one. It's a power rule and our end points air still to the three. The last thing to do is you just plug in the Zen point. So let's start with three first to a natural lot of eight minus two natural log of two and then plus one over three minus one. So that's one half. That's for plugging in three. Now we plug into we have natural log five minus two natural log of one plus one over to minus one, which is just one Now. We simplified combined a CZ, much as we can. We know naturalized of one zero. So let's ignore that. Sir, we have a positive one half minus one so we can take out a negative one half. We have one third natural log of eight. Weaken. Right. This is natural log of eight of the one third if we like, and then we have minus two national law, too, and then we could go ahead and write. This is natural log of five to the one third power. Eight of the one third is just too. So we have natural log of two minus two times naturalized, too. So that gives us the negative one half. So this right here, it's just Ellen too. And it combined these three terms. That's a minus. Ellen, too. And I still have a minus Ellen off. Five to the wonder. So we just plot the minus. Now we could use our lot properties. This is minus one half, minus natural log two times five to the one third, and that's your final answer.

Integration Techniques