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Evaluate the integral. $ \displaystyle \int_2^…

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Problem 19 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_0^1 \frac{x^2 + x + 1}{(x + 1)^2(x + 2)}\ dx $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

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Video Transcript

let's evaluate the definite integral from zero to one, so the denominator is already factored. Let's go ahead and do a partial fraction to composition, so we have a over X plus one. But then, since this is a repeated factor, we have another constant being over X plus one square. And then we have another Lennier factor over here that's different. So that scene over X Plus two and now it's good and multiply by the denominator X plus one squared X plus two. So both sides of this equation So on the left hand side, we're just left with the numerator. But on the right hand side, we have a X plus one times X plus two plus B Times X plus two plus C times. Extras once where. Now let's just go ahead and expand the right hand side. So next day we have X squared plus three X plus two plus B X plus two and then plus E and then X Square plus two X plus one. And now let's go ahead and factor out X Square. That leaves us with a policy, then must pull on X left over with free, eh? A plus B plus two c. And then finally, for the constant, sir, we have to, eh? Plus to be and then plus he So now we compare the coefficients on each side on the left hand side in front of the X squared. We have a one words in the right hand side. We have a policy, so we have a place. He is one. Similarly, the coefficient on the left hand side in front of the ex is the one. The coefficient on the right hand side is three plus B minus to see. So those air people and then finally the constant term on the left is one. The constant term on the right is to a plus two b plus c. So those have to be evil will have a three by three system and the coefficients and BNC, which we can solve. Let's go to the next page. To do this we have a policy was one three eight plus B plus two C equals one, and then we also have to a plus to me, plus C equals one many ways to solve this three by three system. For example, we could take the first equation soft received. And then we could plug this value of see into the other two equations. So plugging that this see value into this equation up here we have three plus me now to see becomes to minus two, eh? And then we could simplify that to become a plus, being equals minus one. Similarly, plug and see into the final equation you have to, eh? Plus to me, plus C, which we write is one minus a that equals one. Now simplify this A plus two b equals zero. So now we have a two by two system that we can solve in and be what? From this last equation here we could saw for a and then we could plug this a value. And so the other equation we'll have negative to be a plus. B equals minus one, which weaken right, solve this B equals one by this equation, up here is negative two times one and then by the first equation over here we have C equals one minus a. So c equals three. Now that we have the values for a B and C, we could plug those into our partial fraction decomposition on page one, So it's going to page three to write this out. We now have the integral from zero to one. Then we have our A over X plus one B, which was one over X plus one squared and then see was three. So we had three over X Plus two, and this is a much easier expression to integrate the problem the way that it was written. And if you don't like the's plus ones and plus twos on the X, you can go out and do a use up for this first one. You could try you equals X plus one for the last one. You can try U equals X Plus two. And even for this one, you could try. U equals X plus one as well. So when we go ahead and integrate these for the first one, we have negative to natural log absolute value at plus one. Then, using the power rule, we have negative one over X plus one and then plus three natural log absolute value X plus two in the end, points zero and one still. Now we just go ahead and plug in one and zero for X, so this is what we get, what we plug in one. Now go ahead and plug in zero. And that's what we have. And now we simplify as much as we can. We have a natural log of one that's zero. And then we'LL combine like Serbs here. So we have negative to natural log to and then minus three natural log to that becomes negative. Five natural log too. And then we have a negative one half. But then we have a plus one. So that's a plus one half and then we have a three ln three left over, and now we could just go ahead and simplify this Me. This is a number of the miracle answer so we could stop here. But if you look to simplify, we can write. This is ln three to the third power. Then I have Ah, let me write this over here is minus ln to the fifth Power. So that's thirty two plus one half. So I just used the law and property. I'm using a lot of property that a natural log B is Ellen B to the power and finally we have a log minus a log. So we can rank. This is a log of a fraction using another log property plus one half, and there's our final answer.

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Problem

Evaluate the integral. $ \displaystyle \int_2^…

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Calculus: Early Transcendentals

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Kristen Karbon

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Kristen Karbon

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals