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Evaluate the integral.

$ \displaystyle \int_{-1}^0 \frac{x^3 - 4x + 1}{x^2 - 3x + 2}\ dx $

$$\frac{5}{2}-\ln 3-\ln 2 \quad \text { or } \quad \frac{5}{2}-\ln 6$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

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Let's evaluate the definite integral from negative one to zero and r inte grin. We see this fraction polynomial divided by polynomial, since the numerator has degree larger than or equal to the denominator. Let's go ahead and do polynomial division. So X cubed, divided by X squared is just X couldn't multiply x by this polynomial on the left and then subtract the whole thing. Three X square minus six X plus one and then three X squared over X squared is just three. So we have three X square minus nine X plus six subtract and we have three X minus fires. So what this tells us is that we can take this in a grand here and write this so we're not looking at the integral just a fraction on the inside. Our question was X plus three. That's up here. Oh, plus, yeah. Can we have a fraction on top? Goes the remainder three x minus five we got from down here are remainder and then you take the original fraction original denominator. And that goes back down here so we can integrate this thing instead. The first term at plus three. That's no fraction there so no partial fraction to composition as the Siri go For the second term, we still have polynomial divided by a polynomial. No. So we should see if that denominator factors and we see it does. We can write it as X minus two X minus one. So this means that we should take three X minus five. This is what the book would call case one for partial freshens. You have distinct linear factors in the denominator, and they're not repeated. So we have a over X minus two and then be over X minus one and now we'LL solve for Andy. It was good and multiply both sides by the denominator on the list. So go ahead and multiply this both sides of this equation here on the left, which is kid three X minus five On the right, we get a X minus one and then be explain, ASU, Let's go ahead and rewrite that right hand side. Let's factor on X. We have a plus me and then we have minus a minus two b. So now we can look at the corresponding coefficients on the left hand side, the number in front of the ex is a three on the right hand side is a plus B, so that gives us one equation. A plus B equals three, and then the constants room on the left is minus five, and on the right hand side, it's minus a minus two B and let's go to the next page. We're a plus. B equals three, and we also have a plus two. B equals five. If we take that second equation from the previous patient, multiply both sides by negative one. We get a plus two b equals five two by two system here for any meet us off. So let's just go ahead and solve this for and be many ways to do this. If we just take the first equation, solve it for a and then plug this a value into the second equation, we get three minus B that's R A plus to be equals five and then just simplify this and we have B equals two and then from this equation over here, plugging this into a We have a equals three minus two, which is one. So that resolves the question from earlier. So we had Insel girl, actually, let's not write the integral yet the previous expression that we had after we do long division we were at this step. Then we did partial fraction the composition for this fraction. So at first three stays the same. And then we had a which was one. So it's a one over X minus two plus B, which was to over X minus one. So now this is the term that will integrate. I'm running out of room here. Let me go to the next page. Negative one zero X plus three one over X minus two two over X minus one. Now, these last two terms are bothering you Because of the subtraction. You can go ahead and do a U substitution. For each of these here, you would do U equals X minus one. When we integrate, we should get X squared over two. That's the power rule three times X. That's just a constant three. So three X natural log Absolute value. X minus two plus two Natural log. Absolute value X minus one and then our end points for a minus. One is zero. Now we just go ahead and plug in the end points we plug in zero the first term in zero, the seconds from a zero. And then we had natural log, absolute value of negative, too. So let's just Ellen too. And when we plug it zero into the last term, we're natural. Either one, which is zero. So we changes. Have Ellen of two for zero, and then we plug in negative one. So negative one squared is just one half, three times minus one. So's to minus three. Plus, then we have natural log negative one minus to an absolute value. So that becomes a three plus two times natural log. Similarly, this becomes too so two times naturalized, too. And the last step here is just to cancel out. But simplify as much as we can. We have one actual longer two and then we're subtracting two of those. So that leaves us with negative natural law too. And then here we have one half minus three. It's negative. Five halfs. We have another negative over here, so it becomes a plus. I have. And then we also have this negative. Ln three here So sloppy on your Asis, Ellen. Three. And we could either stop right there or if you want to simplify this a little more. You could combine those logs. You have a minus. And then we have Ellen to plus Ellen three. So that's using your lawn properties you could write. This is minus ln of two times three, which is Ellen six. So here, either answer should be fine. This or this, and that's your final answer.

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