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