## www.mathcentre.ac.uk/.../9.6a%20Integrating%20Algebraic%20Fractions%20Part2.mp4

• 0:00 - 0:04
In the first of the units on
algebraic fractions, we looked
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at what happened when we had a
proper fraction with linear
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factors in the denominator of
proper fraction with repeated
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linear factors in the
denominator, and what happened
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what I want to do in this video
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is look at what happens when we
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factor when we get an
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will end up with an integral of
something which looks like this
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X plus B.
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Over a X squared plus BX
plus, see where the A&B are
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known constants. And this
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cannot be factorized. Now
there's various things that
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could happen. It's possible that
a could turn out to be 0. Now,
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if it turns out to be 0, what
would be left with?
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Is trying to integrate a
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• 0:58 - 1:04
factor. So we'll just end up
with a B over AX squared plus BX
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plus C. Now the first example
I'd like to show you is what
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we do when we get a situation
where we've just got a
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constant on its own, no ex
terms over the irreducible
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have a look at a specific
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example.
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Suppose we want to
integrate a constant one.
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Over X squared.
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Plus X plus one. We want
to integrate this with
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respect to X.
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This denominator will not
factorize if it would factorize,
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would be back to expressing it
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in partial fractions. The way we
proceed is to try to complete
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the square in the denominator.
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Let me remind you of how we
complete the square for X
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squared plus X plus one.
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It's a complete the square we
try to write the first 2 terms.
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As something squared.
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Well, what do we write in
this bracket? We want an X
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and clearly when the brackets
are all squared out, will get
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an X squared which is that
term dealt with.
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To get an ex here, we need
actually a term 1/2 here because
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you imagine when you square the
brackets out you'll get a half X
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in another half X, which is the
whole X which is that.
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We get something we don't want
when these brackets are all
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squared out, we'll end it with
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1/2 squared. Which is 1/4 and we
don't want a quarter, so I'm
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going to subtract it again here.
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So altogether, all those
terms written down there
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are equivalent to the
first 2 terms over here.
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And to make these equal, we
still need the plus one.
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So tidying this up, we've
actually got X plus 1/2 all
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squared, and one subtract 1/4 is
3/4. That is the process of
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completing the square.
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OK, how will that help us? Well,
it means that what we want to do
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now is considered instead of the
integral we started with. We
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want to consider this integral
one over X plus 1/2 all squared.
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Plus 3/4 we want to integrate
that with respect to X.
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Now, the way I'm going to
proceed is going to make a
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substitution in. Here, I'm going
to let you be X plus 1/2.
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When we do that, are integral
will become the integral of one
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over X plus 1/2 will be just
you, so will end up with you
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squared. We've got plus 3/4.
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We need to take care of the DX.
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Now remember that if we want the
differential du, that's du DX
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DX. But in this case du DX is
just one. This is just one. So
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do you is just DX. So this is
nice and simple. The DX we have
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here just becomes a du.
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Now this integral is a standard
form. There's a standard result
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which says that if you want to
integrate one over a squared
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plus X squared with respect to
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X. That's equal to one over a
inverse tangent that's 10 to the
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minus one of X over a plus a
constant. Now we will use that
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result to write the answer down
to this integral, because this
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is one of these where a.
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Is the square root of 3 over 2?
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That's a squared is 3/4.
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So A is the square root of 3 /
2, so we can write down the
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answer to this straight away and
this will workout at one over a,
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which is one over root 3 over
210 to the minus one.
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Of X over A. In this
case it will be U over a
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which is Route 3 over 2.
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Plus a constant.
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Just to tidy this up a little
bit where dividing by a fraction
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here. So dividing by Route 3
over 2 is like multiplying by
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two over Route 3.
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We've attempted the minus one.
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You we can replace
with X plus 1/2.
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And again, dividing by Route 3
over 2 is like multiplying by
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two over Route 3.
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And we have a constant
at the end.
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And that's the answer. So In
other words, to integrate.
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A constant over an
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We can complete the square as
we did here and then use
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integration by substitution
to finish the problem off.
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So that's what happens when
we get a constant over the
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What else could happen? It may
happen that we get a situation
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like this. We end up with a
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and it's derivative at the top.
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If that happens, it's very
straightforward to finish the
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integration of because we know
from a standard result that this
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evaluates to the logarithm of
the modulus of the denominator
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plus a constant. So, for
example, I'm thinking now of an
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example like this one.
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factor in the denominator.
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Attorney X plus be constant
times X plus another
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constant on the top and if
you inspect this carefully,
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if you look at the bottom
here and you differentiate
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it, you'll get 2X plus one.
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So we've got a situation where
we've got a function at the
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bottom and it's derivative at
the top so we can write this
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is going to be the natural
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logarithm of the modulus of
what's at the bottom.
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Let's see and that's finished.
That's nice and straightforward.
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If you get a situation where
you've got something times X
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plus another constant.
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And this top line is not the
derivative of the bottom
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line. Then you gotta do a bit
more work on it as we'll see
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in the next example.
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Let's have a look at this
example. Suppose we want to
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integrate X divided by X squared
plus X Plus One, and we want to
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integrate it with respect to X.
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Still, if we differentiate, the
bottom line will get 2X.
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Plus One, and that's not
what we have at the top.
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However, what we can do is
we can introduce it to at
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the top, so we have two X in
this following way. By
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little trick we can put a
two at the top.
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And in order to make this the
same as the integral that we
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started with, I'm going to put a
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factor of 1/2 outside. Half and
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the two canceling. Will will
leave the integral that we
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started with that.
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Now. If we differentiate the
bottom you see, we get.
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2X. Which is what we've got
at the top. But we also get a
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plus one from differentiating
the extreme and we haven't
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got a plus one there, so we
apply another little trick
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now, and we do the following.
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We'd like a plus one there.
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So that the derivative of
the denominator occurs in
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the numerator.
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But this is no longer the same
as that because I've added a one
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here. So I've got to take it
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away again. In order that
were still with the same
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problem that we started with.
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Now what I can do is I can split
this into two integrals. I've
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got a half the integral of these
first 2 terms.
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Over X squared plus X plus one.
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DX and I've got a half.
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The integral of the second term,
which is minus one over X
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squared plus X plus one DX so
that little bit of trickery has
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allowed me to split the thing
into two integrals. Now this
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first one we've already seen is
straightforward to finish off,
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because the numerator now is the
derivative of the denominator,
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so this is just a half the
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natural logarithm. Of the
modulus of X squared plus
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X plus one.
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And then we've got minus 1/2.
Take the minus sign out minus
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1/2, and this integral integral
of one over X squared plus X
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Plus one is the one that we did
right at the very beginning.
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And if we just look back, let's
see the results of finding that
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over Route 3 inverse tan of
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twice X plus 1/2 over Route 3?
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So we've got two over Route
3 inverse tangent.
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Twice X plus 1/2.
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Over Route 3.
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Plus a constant of integration.
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I can just tidy this up so it's
nice and neat to finish it off a
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half the logarithm of X squared
plus X plus one.
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The Twos will counsel here, so
I'm left with minus one over
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Route 3 inverse tangent.
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And it might be nice just to
multiply these brackets out to
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finish it off, so I'll have two
X and 2 * 1/2.
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Is one.
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All over Route 3.
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Plus a constant of integration.
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And that's the problem solved.
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Let's have a look at one
final example where we can
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together. Supposing we want
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to integrate 1 divided by XX
squared plus one DX.
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What if we got? In this case,
it's a proper fraction.
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And we've got a linear factor
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factor here.
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You can try, but you'll find
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not factorize, so this is an
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So what we're going to do is
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we're going to, first of all,
express the integrand.
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As the sum of its partial
fractions and the appropriate
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form of partial fractions are
going to be a constant over
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the linear factor.
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And then we'll need BX plus C.
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factor X squared plus one.
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We now have to find abian. See,
we do that in the usual way by
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common denominator will be XX
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squared plus one.
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Will need to multiply top and
bottom here by X squared plus
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one to achieve the correct
denominator so we'll have an AX
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squared plus one.
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And we need to multiply top and
bottom here by X to achieve that
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denominator. So we'll have VX
plus C4 multiplied by X.
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This quantity is equal to that
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quantity. The denominators are
already the same, so we can
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equate the numerators. If we
just look at the numerators
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will have one is equal to a X
squared plus one.
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Plus BX Plus C.
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Multiplied by X.
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What's a sensible value to
substitute for X so we can
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find abian? See while a
sensible value is clearly X is
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zero, whi is that sensible?
Well, if X is zero, both of
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these terms at the end will
disappear.
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So X being zero will have one is
equal to A.
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0 squared is 0 + 1.
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Is still one, so we'll have one
a. So a is one.
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That's our value for a.
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What can we do to find B and see
what I'm going to do now is I'm
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going to equate some
coefficients and let's start by
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looking at the coefficients of X
squared on both side.
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On the left hand side there
are no X squared's.
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hand side? There's clearly
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AX squared here.
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And when we multiply the
brackets out here, that
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would be X squared.
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There are no more X squares, so
A plus B must be zero. That
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means that B must be the minus
negative of a must be the minor
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say, but a is already one, so be
must be minus one.
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We still need to find C and
will do that by equating
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coefficients of X.
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There are no ex terms on
the left.
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There are no ex terms in here.
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There's an X squared term there,
and the only ex term is CX, so
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see must be 0.
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So when we express this in its
partial fractions, will end up
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with a being one.
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Be being minus one
and see being 0.
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So we'll be left with trying to
integrate one over X minus X
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over X squared plus one.
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DX
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so we've used partial fractions
to split this up into two terms,
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and all we have to do now is
completely integration. Let me
• 15:29 - 15:30
write that down again.
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We want to integrate one over X
minus X over X squared plus one
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and all that wants to be
integrated with respect to X.
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First term straightforward.
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The integral of one
over X is the
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logarithm of the
modulus of X.
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To integrate the second term.
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We notice that the numerator is
almost the derivative of the
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denominator. If we
differentiate, the denominator
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will get 2X.
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There is really only want 1X
now. We fiddle that by putting
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it to at the top and a half
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outside like that. So this
integral is going to workout
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to be minus 1/2 the
logarithm of the modulus of
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X squared plus one.
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And there's a constant of
integration at the end.
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And I'll leave the answer like
that if you wanted to do. We
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could combine these using the
laws of logarithms.
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And that's integration of
algebraic fractions. You
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need a lot of practice at
that, and there are more
• 16:41 - 16:43
practice exercises in the
accompanying text.
Title:
www.mathcentre.ac.uk/.../9.6a%20Integrating%20Algebraic%20Fractions%20Part2.mp4
Video Language:
English