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Evaluate the integral.

$ \displaystyle \int_1^2 \frac{x^3 + 4x^2 + x - 1}{x^3 + x^2}\ dx $

$$\int_{1}^{2} \frac{x^{3}+4 x^{2}+x-1}{x^{3}+x^{2}} d x=\ln 6+\frac{1}{2}$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Harvey Mudd College

Baylor University

University of Nottingham

Idaho State University

Lectures

01:11

In mathematics, integratio…

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Evaluate the indefinite in…

08:52

Let's evaluate the definite integral from once it's who we see. The answer brand. We have polynomial divided by polynomial, since the numerator and degree have the same denominator. Let's go ahead and do a polynomial division now x cubed, divided by excuses, just one. So we have X cubed plus X where cigarette and we have four x squared, minus X square, three X squared plus X and then minus one. And this is our remainder because X cubed over here does not go into exploiter. So this means we can write the fraction as our quotient, which was one then we had a remainder of three x squared plus X minus one, and then we have the original denominator on the bottom. So now we're ready to do partial fraction to composition. And with first thing is to go to that denominator and then factor out etzler. And then we have X squared. Plus one hopes X plus one left over. This is what the book would call case, too, because we have a repeated factor here we have X bait was written as a parent twice. In other words, I'm just saying X squared is X times X So this is repeated. So using with the book calls case to here, we can rewrite this fraction. So first, let me write this out Then I have X squared X plus one. So case, too, as the author cause it we'LL have Rex be over ex player and then another constant over X plus one. So the next step here, which will do on the next page, is to go ahead and multiply both sides by the denominator on the left. All right, answer the next page after multiplying denominated under left gets cancelled out. Then we have a X Times that's plus one B plus one and then see X where Let's go ahead and just take this right hand side and was first, Let's rewrite the left and the right hand side. Was he? We can rewrite. This is X squared already got a room up there That's just X squared plus X. So it's factor on X squared. We're a and then we also have seen. Then let's fact, throughout the ex term, we'LL have a again and then we'LL have a B is well and then for the constant term, we'LL just have a B So looking at the coefficient on the left and the right in front of the X squared, we see a three and on the right a placide, I guess This one equation in front of the X on the left which you see a one on the right, A plus B Hey, that gives you another equation. And then finally under left, we see negative one on the right. We see a beat for the constant term. So, like, this is B equals negative one. And that's already one of our values. Let's plug this into the previous equation and green. So plug in B equals minus one. You have a plus Negative one equals one for just a equals two and then plug this value of A into the equation and red, you have two plus C equals three, so C equals one. So now we have the integral let's actually go to the next page. So we'LL just plug in our values for a B and C into the partial fraction the composition on the previous page and we'LL immigrate. So we have one plus and then actually here, let me take a step back just in case, we had forgotten what's on the first page. So we were at this step right here after we did long division. This was the thing that we were in her reading. And then we did the parcel Fraction the composition. And we just wrote This is now we have two of Rex negative one over X Claire, because be was minus one evidence he was one. So one over X plus one. And now the next step is just to integrate. So she's got an ingrate, were integrating from one to two once a two. Now we can go ahead and integrate these. So the first term just becomes an ex to natural lot of X absolute value in there, too. And then using the power rule will have a one over X. And then finally, this looks very similar to the second integral. So if that plus one is bothering you, just go ahead and do a u substitution. And after doing so, you should get natural log Absolute value, X plus one, and we have our own points. Wants it too. Now we just plug these in points in plug in the two first. That's what we put into. And then when I plug in one one plus two Ellen one, which would be zero plus one over one plus Eleanor, too. Now we just cancel oppa's much as we can so weigh. See that this positives? Who here? We'll cancel with this negative two on the bottom. So we have a one half left over, and then we have a two natural law to minus natural on tube. So luscious natural lot of two and then we have a plus ln three so we can write. That is our answer. Or we could use a property of the logs to write. This is Ellen of two times three. So it's Ellen of six. Either one of these is fun, and that's your final answer.

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