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