WEBVTT

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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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when we had improper fractions.
what I want to do in this video

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is look at what happens when we
get an irreducible quadratic

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factor when we get an
irreducible quadratic factor

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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
quadratic in the denominator

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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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constant. Over this quadratic

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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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quadratic factor, so let's
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
irreducible quadratic factor.

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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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quadratic factor.

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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
quadratic function at the bottom

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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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Again, irreducible quadratic
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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down straight away. The answer
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

00:09:04.120 --> 00:09:04.880
the numerator.

00:09:05.960 --> 00:09:08.914
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

00:09:13.440 --> 00:09:14.540
problem that we started with.

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Now what I can do is I can split
this into two integrals. I've

00:09:21.258 --> 00:09:24.628
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

00:09:45.020 --> 00:09:49.068
allowed me to split the thing
into two integrals. Now this

00:09:49.068 --> 00:09:52.748
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

00:10:10.076 --> 00:10:13.532
1/2, and this integral integral
of one over X squared plus X

00:10:13.532 --> 00:10:17.276
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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integral. Was this one here, two
over Route 3 inverse tan of

00:10:26.475 --> 00:10:28.890
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

00:10:49.340 --> 00:10:52.340
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.

00:11:09.110 --> 00:11:09.820
Is one.

00:11:10.890 --> 00:11:12.178
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

00:11:30.280 --> 00:11:33.700
draw some of these threads
together. Supposing we want

00:11:33.700 --> 00:11:37.500
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.

00:11:44.030 --> 00:11:45.482
And we've got a linear factor

00:11:45.482 --> 00:11:47.880
here. And a quadratic
factor here.

00:11:48.900 --> 00:11:52.376
You can try, but you'll find
that this quadratic factor will

00:11:52.376 --> 00:11:54.272
not factorize, so this is an

00:11:54.272 --> 00:11:57.754
irreducible quadratic factor.
So what we're going to do is

00:11:57.754 --> 00:11:59.932
we're going to, first of all,
express the integrand.

00:12:02.580 --> 00:12:06.050
As the sum of its partial
fractions and the appropriate

00:12:06.050 --> 00:12:09.867
form of partial fractions are
going to be a constant over

00:12:09.867 --> 00:12:10.908
the linear factor.

00:12:12.870 --> 00:12:15.586
And then we'll need BX plus C.

00:12:16.850 --> 00:12:20.378
Over the irreducible quadratic
factor X squared plus one.

00:12:22.450 --> 00:12:28.255
We now have to find abian. See,
we do that in the usual way by

00:12:28.255 --> 00:12:31.738
adding these together, the
common denominator will be XX

00:12:31.738 --> 00:12:32.899
squared plus one.

00:12:35.200 --> 00:12:39.208
Will need to multiply top and
bottom here by X squared plus

00:12:39.208 --> 00:12:42.882
one to achieve the correct
denominator so we'll have an AX

00:12:42.882 --> 00:12:43.884
squared plus one.

00:12:45.530 --> 00:12:51.004
And we need to multiply top and
bottom here by X to achieve that

00:12:51.004 --> 00:12:54.914
denominator. So we'll have VX
plus C4 multiplied by X.

00:12:57.330 --> 00:12:59.160
This quantity is equal to that

00:12:59.160 --> 00:13:03.143
quantity. The denominators are
already the same, so we can

00:13:03.143 --> 00:13:07.303
equate the numerators. If we
just look at the numerators

00:13:07.303 --> 00:13:11.252
will have one is equal to a X
squared plus one.

00:13:13.650 --> 00:13:15.518
Plus BX Plus C.

00:13:16.820 --> 00:13:18.410
Multiplied by X.

00:13:20.400 --> 00:13:23.777
What's a sensible value to
substitute for X so we can

00:13:23.777 --> 00:13:27.154
find abian? See while a
sensible value is clearly X is

00:13:27.154 --> 00:13:30.838
zero, whi is that sensible?
Well, if X is zero, both of

00:13:30.838 --> 00:13:32.987
these terms at the end will
disappear.

00:13:34.520 --> 00:13:38.304
So X being zero will have one is
equal to A.

00:13:39.080 --> 00:13:41.078
0 squared is 0 + 1.

00:13:42.080 --> 00:13:45.032
Is still one, so we'll have one
a. So a is one.

00:13:46.830 --> 00:13:48.090
That's our value for a.

00:13:49.410 --> 00:13:54.170
What can we do to find B and see
what I'm going to do now is I'm

00:13:54.170 --> 00:13:56.690
going to equate some
coefficients and let's start by

00:13:56.690 --> 00:13:59.490
looking at the coefficients of X
squared on both side.

00:14:01.560 --> 00:14:04.730
On the left hand side there
are no X squared's.

00:14:06.350 --> 00:14:09.626
What about on the right
hand side? There's clearly

00:14:09.626 --> 00:14:10.718
AX squared here.

00:14:12.990 --> 00:14:15.339
And when we multiply the
brackets out here, that

00:14:15.339 --> 00:14:16.383
would be X squared.

00:14:19.960 --> 00:14:24.202
There are no more X squares, so
A plus B must be zero. That

00:14:24.202 --> 00:14:28.444
means that B must be the minus
negative of a must be the minor

00:14:28.444 --> 00:14:32.080
say, but a is already one, so be
must be minus one.

00:14:34.150 --> 00:14:38.518
We still need to find C and
will do that by equating

00:14:38.518 --> 00:14:39.610
coefficients of X.

00:14:41.740 --> 00:14:43.484
There are no ex terms on
the left.

00:14:45.240 --> 00:14:46.780
There are no ex terms in here.

00:14:48.460 --> 00:14:53.248
There's an X squared term there,
and the only ex term is CX, so

00:14:53.248 --> 00:14:54.616
see must be 0.

00:14:56.310 --> 00:14:59.946
So when we express this in its
partial fractions, will end up

00:14:59.946 --> 00:15:01.158
with a being one.

00:15:02.720 --> 00:15:06.704
Be being minus one
and see being 0.

00:15:09.020 --> 00:15:14.987
So we'll be left with trying to
integrate one over X minus X

00:15:14.987 --> 00:15:17.282
over X squared plus one.

00:15:18.990 --> 00:15:19.640
DX

00:15:21.610 --> 00:15:25.294
so we've used partial fractions
to split this up into two terms,

00:15:25.294 --> 00:15:28.978
and all we have to do now is
completely integration. Let me

00:15:28.978 --> 00:15:30.206
write that down again.

00:15:31.750 --> 00:15:38.386
We want to integrate one over X
minus X over X squared plus one

00:15:38.386 --> 00:15:43.600
and all that wants to be
integrated with respect to X.

00:15:46.660 --> 00:15:48.430
First term straightforward.

00:15:49.260 --> 00:15:52.388
The integral of one
over X is the

00:15:52.388 --> 00:15:54.734
logarithm of the
modulus of X.

00:15:56.570 --> 00:15:58.320
To integrate the second term.

00:15:59.040 --> 00:16:03.506
We notice that the numerator is
almost the derivative of the

00:16:03.506 --> 00:16:05.942
denominator. If we
differentiate, the denominator

00:16:05.942 --> 00:16:07.160
will get 2X.

00:16:08.110 --> 00:16:11.230
There is really only want 1X
now. We fiddle that by putting

00:16:11.230 --> 00:16:13.310
it to at the top and a half

00:16:13.310 --> 00:16:17.598
outside like that. So this
integral is going to workout

00:16:17.598 --> 00:16:21.538
to be minus 1/2 the
logarithm of the modulus of

00:16:21.538 --> 00:16:23.114
X squared plus one.

00:16:24.710 --> 00:16:27.356
And there's a constant of
integration at the end.

00:16:29.400 --> 00:16:32.611
And I'll leave the answer like
that if you wanted to do. We

00:16:32.611 --> 00:16:34.587
could combine these using the
laws of logarithms.

00:16:35.670 --> 00:16:37.861
And that's integration of
algebraic fractions. You

00:16:37.861 --> 00:16:41.304
need a lot of practice at
that, and there are more

00:16:41.304 --> 00:16:43.182
practice exercises in the
accompanying text.