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

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.