
In this video, we're going to
have a look at how we can

integrate algebraic
fractions. The sorts of

fractions that we're going to
integrate and like these

here.

Now, superficially, they will
look very similar, but there are

important differences which I'd
like to point out when you come

to tackle a problem of
integrating a fraction like

this, it's important that you
can look for certain features,

for example, in this first

example. In the denominator we
have what we call 2 linear

factors. These two linear
factors are two different linear

factors. When I say linear
factors, I mean there's no X

squared, no ex cubes, nothing
like that in it. These are just

of the form a X Plus B linear
factors. A constant times X plus

another constant. So with two
linear factors here.

This example is also got linear
factors in clearly X Plus One is

a linear factor. X minus one is
a linear factor, but the fact

that I've got an X minus one
squared means that we've really

got X minus one times X minus 1
two linear factors within here.

So we call this an example of a
repeated linear factor.

Linear factors, repeated linear
factors. Over here we've got a

quadratic. Now this particular
quadratic will not factorize and

because it won't factorize, we
call it an irreducible quadratic

factor. This final example here
has also got a quadratic factor

in the denominator, but unlike
the previous one, this one will

in fact factorize because X
squared minus four is actually

the difference of two squares
and we can write this as X minus

2X plus two. So whilst it might
have originally looked like a

quadratic factor, it was in fact
to linear factors, so that's

what those are one of the
important things we should be

looking for when we come to
integrate quantities like these.

Weather we've got linear

factors. Repeated linear
factors, irreducible quadratic

factors, or quadratic factors
that will factorize.

Something else which is
important as well, is to examine

the degree of the numerator and
the degree of the denominator in

each of these fractions.
Remember, the degree is the

highest power, so for example in
the denominator of this example

here, if we multiplied it all
out, we actually get the highest

power as three, because when we
multiply the first terms out

will get an X squared, and when
we multiply it with the SEC

bracket, X Plus, one will end up
with an X cubed. So the degree

of the denominator there is 3.

The degree of the numerator is 0
because we can think of this as

One X to the 0.

In the first case, we've got an
X to the one here, so the degree

of the numerator there is one,
and if we multiply the brackets

out, the degree of the of the
denominator will be two. Will

get a quadratic term in here. In
this case, the degree of the

numerator is 0. And in this
case, the degree of the

denominator is to the highest
power is too. So in all of

these cases, the degree of the
numerator is less than the

degree of the denominator, and
we call fractions like these

proper fractions.

On the other hand, if we look at
this final example, the degree

of the numerator is 3, whereas
the degree of the denominator is

too. So in this case, the degree
of the numerator is greater than

the degree of the denominator,
and this is what's called an

improper fraction. Now when we
start to integrate quantities

like this will need to examine
whether we're dealing with

proper fractions or improper
fractions, and then, as I said

before, will need to look at all
the factors in the denominator.

Will also need to call appan
techniques in the theory of

partial fractions. There is a
video on partial fractions and

you may wish to refer to that if

necessary. If you have a linear
factor in the denominator, this

will lead to a partial fraction
of this form. A constant over

the linear factor.

If you have a repeated linear
factor in the denominator,

you'll need two partial
fractions. A constant over

the factor and a constant
over the factor squared.

Finally, if you have a quadratic
factor which is irreducible,

you'll need to write a partial
fraction of the form a constant

times X plus another constant
over the irreducible quadratic

factor, so will certainly be
calling upon the techniques of

partial fractions. Will also
need to call appan. Lots of

techniques and integration.

I'm just going to mention just
two or three here, which will

need to use as we proceed
through the examples of 1st.

Crucial result is the standard
result, which says that if you

have an integral consisting of a
function in the denominator.

And it's derivative in the
numerator. Then the result is

the logarithm of the modulus of
the function in the denominator.

So for example, if I ask you to
integrate one over X plus one

with respect to X.

Then clearly the function in the
denominator is X plus one.

And its derivative is one which
appears in the numerator. So

we've an example of this form.

So the resulting integral is the
logarithm of the modulus of the

function that was in the
denominator, which is X plus

one. Plus a constant of
integration, so we will need

that result very frequently in
the examples which are going to

follow will also need some
standard results and one of the

standard results I will call
appan is this one. The integral

of one over a squared plus X
squared is one over a inverse

tan of X over a plus C.

Results like this can be found
in tables of standard integrals.

Finally, we need to integrate
quantities like this and you'll

need to do this probably using
integration by substitution. An

integral like this can be worked
out by making the substitution

you equals X minus one.

So that the differential du
is du DX.

DX, which in this case is du
DX, will be just one.

So do you is DX and that's
integral. Then will become

the integral of one over
you, squared du.

One over you squared is the same
as the integral of you to the

minus 2D U, which you can solve
by integrating increasing the

power BI want to give you you to
the minus one over minus one.

Plus a constant of integration.

This can be finished off by
changing the you back to the

original variable X minus one
and that will give us X minus

one to the minus one over minus

one plus C. Which is the same as
minus one over X minus one plus

C, which is the results I have

here. So what I'm saying is that
throughout the rest of this unit

will need to call Appan lots of
different techniques to be able

to perform the integrals.

As we shall see.

Let's look at the

first example. Suppose we
want to integrate this algebraic

fraction. 6 / 2

minus X. X

+3
DX

The first thing we do is we look
at the object we've got and try

to ask ourselves, are we dealing
with a proper or improper

fraction and what are the
factors in the denominator like?

Well, if we multiply the power,
the brackets at the bottom will

find that the highest power of X
is X squared, so the degree of

the denominator is 2.

The highest power in the
numerator is one. This is an X

to the power one, so the degree
of the numerator is one because

the degree of the numerator is
less than the degree of the

denominator. This is an example
of a proper fraction.

Both of these factors
in the denominator.

Are linear factors.

So we're dealing with a proper
fraction with linear factors.

The way we proceed is to take
this fraction and express it in

partial fractions. So I'll start
with the fraction again.

And express it in the
appropriate form of partial

fractions. Now because it's

proper. And because we've got
linear factors, the appropriate

form is to have a constant over
the first linear factor.

Plus another constant over the
second linear factor.

Our task now is to find values
for the constants A&B.

Once we've done that, will
be able to evaluate this

integral by evaluating these
two separately.

So to find A and be the first
thing we do is we add these

two fractions together again.

Remember that to add 2 fractions
together, we've got to give them

the same denominator. They've
got to have a common

denominator. The common
denominator is going to be made

up of the two factors. 2 minus

X&X +3. To write the first term
as an equivalent fraction with

this denominator, we multiply
top and bottom by X plus three.

So if we multiply top here by X
+3 and bottom there by X +3.

Will achieve this fraction.

And this fraction is equivalent
to the original 1.

Similarly with the second term.

To achieve a common denominator
of 2 minus XX +3.

I need to multiply top and

bottom here. By two minus X, so
B times 2 minus X and this

denominator times 2 minus X and
that will give me.

B2 minus X at the top.

Now these two fractions
have the same denominator,

we can add them together
simply by adding the

numerators together, which
will give us a multiplied

by X +3.

Plus B multiplied by
two minus X.

All divided by the common

denominator. What we're saying
is that this fraction we

started with is exactly the

same. As this quantity here.

Now the denominators
are already the same.

So if this is the same as that,
and the denominators are already

the same, then so too must be
the numerators, so we can equate

the numerators if we equate the
numerators we can write down X

equals. AX
+3.

Plus B2 Minus
X.

This is the equation that's
going to allow us to calculate

values for A&B.

Now we can find values for A&B
in one of two different ways.

The 1st way that I'm going to

look at. Is to substitute
specific values in for X.

Remember that this quantity on
the left is supposed to be equal

to this on the right for any
value of X at all. So in

particular, we can choose any
values that we like. That will

make all this look simpler.

And what I'm going to do is I'm
going to choose X to have the

value to. Why would I do that?

I choose X to have the value
too, because then this second

term will become zero and have 2
 2, which is zero. Will lose

this term. And we'll be able to

calculate A. So by careful
choice of values for X, we can

make this look a lot simpler.

So with X is 2.

On the left will have two.

On the right will have 2 +
3, which is 5 times a.

And this term will vanish.

This gives me a value for a
straightaway dividing both sides

by 5. I can write that a is 2/5.

We need to find B.

Now a sensible value that will
enable us to find B is to let X

be minus three whi, is that?

Well, if X is minus three,
will have minus 3 + 3, which

is zero. And all of this
first term will vanish.

And we'll be able to find be so
letting XP minus three will have

minus three on the left zero
from this term here, and two

minus minus three, which is 2 +
3, which is 55B.

Dividing both sides by 5 will
give us a value for B's

minus three over 5.

So now we know a value for a. We
know a value for B.

And we can then proceed to
evaluate the integral by

evaluating each of these

separately. Let me write
this down again. We want

the integral of X divided
by 2 minus XX +3.

With respect to X.

We expressed this algebraic
fraction in its partial fraction

in its partial fractions, and we
found that a was 2/5.

And be was

minus 3/5. So instead
of integrating this original

fraction, what we're going to do
now is integrate separately the

two partial fractions.

And will integrate these
separately and will do it like

this. In the first integral,
we're going to take out the

factor of 2/5. I will be left
with the problem of integrating

one over 2 minus X with respect

to X. For the second,
we're going to take out

minus 3/5. And integrate
one over X +3 with

respect to X.

So the problem of
integrating this algebraic

fraction has been split
into the problem of

evaluating these two
separate integrals and both

of these are simpler than
the one we started with.

Let's deal with the second one
first. The second one is a

situation where we've got a
function at the bottom and it's

derivative at the top. Because
we've got X plus three at the

bottom and the derivative of X
+3 is just one which appears at

the top. So this just evaluates
to minus 3/5 the natural

logarithm of the modulus of
what's at the bottom.

We've got the similar situation
here, except if you

differentiate the denominator,
you get minus one because of

this minus X, so we'd really
like a minus one at the top.

And I can adjust my numerator to
make it minus one, provided that

I counteract that with putting a
minus sign outside there.

So we can write all this as
minus 2/5 the natural logarithm

of the modulus of 2 minus X. And
of course we need a constant of

integration at the very end.

So that's the result of
integrating X over 2 minus XX +3

and the problems finished.

What I'd like to do is just go
back a page and just show you an

alternative way of calculating
values for the constants A&B in

the partial fractions, and I
want us to return to this

equation here that we use to
find A&B. Let me write that

equation down again.

X is equal to
a X +3.

Plus B2 Minus

X. What I'm going
to do is I'm going to

start by removing the brackets.

Will have a multiplied by X

AX. A Times 3
which is 3A.

B times two or two B.

And be times minus X or
minus BX.

And then what I'm going to do
is, I'm going to collect similar

terms together so you see Ivan
Axe here and minus BX there.

So altogether I have a minus B,
lots of X.

And we've got 3A Plus
2B here.

We now use this equation to
equate coefficients on both

sides. What do we mean by that?
Well, what we do is we ask

ourselves how many X terms do we
have on the left and match that

with the number of X terms that
we have on the right. So you

see, on the left hand side here,
if we look at just the ex terms,

there's 1X. On the right we've
got a minus B, lots of X.

So we've equated the
coefficients of X on both sides.

We can also look at constant
terms on both sides. You see the

three A plus 2B is a constant.

There are no constant terms on
the left, so if we just look at

constants, there are none on the

left. And on the right there's
3A Plus 2B.

And you'll see what we have.
Here are two simultaneous

equations for A&B and if we
solve these equations we can

find values for A&B. Let me call
that equation one and that one

equation two. What I'm going to
do is I'm going to multiply

equation one by two so that will
end up with two be so that we

will be able to add these
together to eliminate the bees.

So if I take equation one and I
multiply it by two, I'll get 2

ones or two. 2A minus 2B.
Let's call that equation 3.

If we add equations two and

three together. We've got 0 + 2,
which is 2, three, 8 + 2 A which

is 5A and two be added to minus
2B cancels out, so two is 5. In

other words, A is 2 over 5,
which is the value we had

earlier on for A.

We can then take this value for
A and substitute it in either of

these equations and obtain a
value for be. So, for example,

if we substitute in the first

equation. Will find that one
equals a, is 2/5 minus B.

Rearranging this B is equal to
2/5  1.

And 2/5  1 is minus 3/5 the
same value as we got before.

So we've seen two ways of
finding the values of the

constants A&B. We can substitute
specific values for X or we can

equate coefficients on both

sides. Often will need to use a
mix of the two methods in order

to find all the constants in a

given problem. Let's have
a look at

a definite integral.

Suppose we want to find the
integral from X is one to access

2. Of three divided by
XX plus one with respect to

X. As before, we examine this
integrand and ask ourselves, is

this a proper or improper

fraction? Well, the degree of
the denominator is too, because

when we multiply this out, the
highest power of X will be 2.

The degree of the numerator is
zero, with really 3X to the zero

here, so this is an example of a

proper fraction. On both of
these factors are linear

factors. So as before, I'm going
to express the integrand.

As the sum of its partial
fractions. So let's do that

first of all. 3 divided by XX

plus one. The appropriate form
of partial fractions.

Are constant. Over the
first linear factor.

Plus another constant over the
second linear factor.

And our job now is to try to
find values for A&B.

We do this by adding these
together as we did before,

common denominator XX plus one
in both cases.

To write a over X as an
equivalent fraction with

this denominator will
need to multiply top and

bottom by X plus one.

To write B over X plus one with
this denominator will need to

multiply top and bottom by X.

So now we've given these two
fractions a common denominator,

and we add the fractions
together by adding the

numerators. I'm putting the
result over the common

denominator. So 3 divided by XX
plus One is equal to all this.

The denominators are already the

same. So we can equate the
numerators that gives us the

equation 3 equals a X plus one
plus BX. And this is the

equation we can use to try to
find values for A&B.

We could equate coefficients, or
we can substitute specific

values for X and what I'm going
to do is I'm going to substitute

the value X is not and the
reason why I'm picking X is not

is because I recognize straight
away that's going to.

Kill off this last term here
that'll have gone and will be

able to just find a value for A.

So we substitute X is not on the
left, will still have 3.

And on the right we've got not
plus one which is one 1A.

Be times not is not so that

goes. So In other words, we've
got a value for A and a is 3.

Another sensible value to
substitute is X equals minus

one. Why is that a sensible

value? Well, that's a sensible
value, because if we put X is

minus one in minus one plus one
is zero and will lose this first

term with the A in and will now

be. So putting X is minus one
will have 3.

This will become zero and will
have be times minus one which is

minus B. So this tells us that B
is actually minus three.

So now we know the value of a is
going to be 3 and B is going to

be minus three. And the problem
of performing this integration

can be solved by integrating
these two terms separately.

Let's do that now. I'll write
these these terms down again.

We're integrating three over XX
plus one with respect to X.

And we've expressed already this
as its partial fractions, and

found that we're integrating
three over X minus three over X,

plus one with respect to X, and
this was a definite integral. It

had limits on, and the limits
will one and two.

So now we use partial
fractions to change this

algebraic fraction into
these two simple integrals.

Now, these are
straightforward to finish

because the integral of
three over X is just three

natural logarithm of the
modulus of X.

The integral of three over X

plus one. Is 3 natural logarithm
of the modulus of X plus one?

And there's a minus sign
in the middle from that.

This is a definite

integral. So we have square
brackets and we write the limits

on the right hand side.

The problem is nearly
finished. All we have to do

is substitute the limits in.

Upper limit first when X is
2, will have three natural

log of two.

When X is 2 in here will have

minus three. Natural
logarithm of 2 + 1, which is

3. So that's what we get when
we put the upper limit in.

When we put the lower limit in,
when X is one will have three

natural logarithm of 1.

Minus three natural logarithm of
1 + 1, which is 2. So that's

what we get when we put the
lower limiting. And of course we

want to find the difference of
these two quantities.

Here you'll notice with three

log 2. And over here
there's another three

log, two with A minus and
minus, so we're adding

another three log 2. So
altogether there will be

6 log 2.

That's minus three log 3.

And the logarithm of

1. Is 0 so that banishes.

Now we could leave the answer
like that, although more often

than not would probably use
the laws of logarithms to try

to tighten this up a little
bit and write it in a

different way. You should be
aware that multiplier outside,

like this six, can be put
inside as a power, so we can

write this as logarithm of 2
to the power 6.

Subtract again a multiplier
outside can move inside as a

power so we can write this as
logarithm of 3 to the power 3.

And you'll also be aware from
your loss of logarithms that if

we're finding the difference of
two logarithms, and we can write

that as the logarithm of 2 to
the power 6 / 3 to the power 3.

And that's my final answer.

Let's look at another example in
which the denominator contains a

repeated linear factor. Suppose
we're interested in evaluating

this integral 1 divided by X
minus one all squared X Plus

One, and we want to integrate
that with respect to X.

So again, we have a proper

fraction. And there is a linear
factor here. X plus one, another

linear factor X minus one. But
this is a repeated linear factor

because it occurs twice.

The appropriate form of partial
fractions will be these.

We want to
constant over the

linear factor X

minus one. We want
another constant over the linear

factor repeated X minus one
squared. And finally we need

another constant. See over this
linear factor X plus one.

And our task is before
is to try to find values

for the constants AB&C.

We do that as before, by
expressing each of these over a

common denominator and the
common denominator that we want

is going to be X minus one.

Squared X plus
one.

Now to achieve a common
denominator of X minus one

squared X Plus one will need to
multiply the top and bottom here

by X minus 1X plus one. So we
have a X minus 1X plus one.

To achieve the common
denominator in this case will

need to multiply top and bottom
by X plus one.

So we'll have a BX plus one.

And finally, in this case, to
achieve a denominator of X minus

one squared X Plus one will need
to multiply top and bottom by X

minus 1 squared.

Now this fraction here
is the same as

this fraction here.

Their denominators are already
the same, so we can equate the

numerators. So if we just look
at the numerators will have one

on the left is equal to the top
line here on the right hand

side. AX minus
1X plus one.

Plus B. X plus one.

Plus C. X minus one all

squared. And now we choose some
sensible values for X, so

there's a lot of these terms
will drop away. For example,

supposing we pick X equals 1,
what's the point of picking X

equals one? Well, if we pick
axes one, this first term

vanishes. We lose a.

Also, if we pick X equal to 1,
the last term vanish is because

we have a 1  1 which is zero
and will be just left with the

term involving be. So by letting
XP, one will have one.

On the left is equal to 0.

One and one here is 22B.

And the last term vanishes. In
other words, B is equal to 1/2.

What's another sensible value to
pick for X? Well, if we let X

equal minus one.

X being minus one will mean that
this term vanish is minus one

plus One is 0.

This term will vanish minus one,
plus one is zero and will be

able to find see. So I'm going
to let X be minus one.

Will still have the one on the

left. When X is minus one, this

goes. This goes.

And on the right hand side this
term here will have minus 1  1

is minus two we square it will
get plus four, so will get plus

4C. In other words, see is going
to be 1/4.

So we've got be. We've got C and
now we need to find a value for

a. Now we can substitute any
other value we like in here, so

I'm actually going to pick X
equals 0. It's a nice simple

value. Effects is zero, will
have one on the left if X is

zero there and there will be
left with minus one.

Minus 1 * 1 is minus 1  1
times a is minus a.

Thanks is 0 here. Will just left
with B Times one which is be.

And if X is 0 here will have
minus one squared, which is plus

one plus One Times C Plus C. Now
we already know values for B and

for C, so we substitute these in
will have one is equal to minus

A plus B which is 1/2 plus C
which is 1/4.

So rearranging this will have
that a is equal to.

Well, with a half and a quarter,
that's 3/4 and the one from this

side over the other side is
minus one. 3/4  1 is going to

be minus 1/4.

So there we have our values for
a for B and for C, so a being

minus 1/4 will go back in there.

Be being a half will go back in
here and see being a quarter

will go back in there and then
we'll have three separate pieces

of integration to do in order to
complete the problem.

Let me write these all down
again. I'm going to write the

integral out and write these
three separate ones down.

So we had that the integral of
one over X minus one squared X

Plus one, all integrated with
respect to X, is going to equal.

Minus 1/4 the integral of one
over X minus one.

They'll be plus 1/2 integral of
one over X minus 1 squared.

And they'll
be 1/4

over X

plus one.
And we want

to DX in

every term. So we've
used partial fractions to split

this integrand into three
separate terms, and will try and

finish this off. Now. The first
integral straightforward when we

integrate one over X minus one
will end up with just the

natural logarithm of the
denominator, so we'll end up

with minus 1/4 natural
logarithm. The modulus of X

minus one. To integrate this
term, we're integrating one over

X minus one squared. Let me just
do that separately.

To integrate one over X minus
one all squared, we make a

substitution and let you equals
X minus one.

Do you then will equal du DX,
which is just one times DX, so

do you will be the same as DX?

The integral will become the
integral of one over.

X minus one squared
will be you squared.

And DX is D.

Now, this is straightforward to
finish because this is

integrating you to the minus 2.

Increase the power by one
becomes you to the minus 1

divided by the new power.

And add a constant of
integration.

So when we do this, integration
will end up with minus one over

you. Which is minus one over.

X minus one.

And we can put that back into
here now. So the integral half

integral of one over X minus one
squared will be 1/2.

All that we've got down here,
which is minus one over X minus

one. The constant of integration
will add another very end. This

integral is going to be 1/4.

The natural logarithm of the
modulus of X Plus One and then a

single constant for all of that.

Let me just tidy up this little
a little bit. The two logarithm

terms can be combined. We've got
a quarter of this logarithm.

Subtract 1/4 of this logarithm,
so together we've got a quarter

the logarithm. If we use the
laws of logarithms, we've gotta

log subtracted log. So we want
the first term X Plus one

divided by the second term X

minus one. And then this term
can be written as minus 1/2.

One over X minus one.

Is a constant of integration
at the end, and that's the

integration complete.

Now let's look at an example in
which we have an improper

fraction. Suppose we have this

integral. The
degree of

the numerator

is 3.
The degree of the

denominator is 2.

3 being greater than two
means that this is an

improper fraction.

Improper fractions requires
special treatment and the first

thing we do is division
polynomial division. Now, if

you're not happy with Long
Division of polynomials, then I

would suggest that you look
again at the video called

polynomial division. When
examples like this are done very

thoroughly, but what we want to
do is see how many times X minus

X squared minus four will divide

into X cubed. The way we do
this polynomial division is we

say how many times does X
squared divided into X cubed.

That's like asking How many
times X squared will go into X

cubed, and clearly when X
squared is divided into X cubed,

the answer is just X.

So X squared goes into X
cubed X times and we write

the solution down there.

We take what we have just
written down and multiply it by

everything here. So X times X
squared is X cubed.

X times minus four is minus 4X.

And then we subtract.

X cubed minus X cubed vanish is.

With no access here, and we're
subtracting minus 4X, which is

like adding 4X.

This means that when we divide X
squared minus four into X cubed,

we get a whole part X and a
remainder 4X. In other words, X

cubed divided by X squared minus
four can be written as X.

Plus 4X divided by X
squared minus 4.

So in order to tackle this
integration, we've done the long

division and we're left with two
separate integrals to workout.

Now this one is going to be
straightforward. Clearly you're

just integrating X, it's going
to be X squared over 2. This is

a bit more problematic. Let's
have a look at this again.

This is now a proper fraction
where the degree of the

numerator is one with the next
to the one here.

The degree of the denominator is

2. So it's a proper fraction.

It looks as though we've got a
quadratic factor in the

denominator. But in fact, X
squared minus four will

factorize. It's in fact, the
difference of two squares. So we

can write 4X over X squared
minus four as 4X over X minus 2X

plus two. So the quadratic will
actually factorize and you'll

see. Now we've got two linear

factors. We can express this in
partial fractions. Let's do

that. We've got 4X over X, minus
two X +2, and because each of

these are linear factors, the
appropriate form is going to be

a over X minus 2.

Plus B over X +2.

We add these together using a
common dumped common denominator

X minus two X +2 and will
get a X +2 plus BX minus

2. All over the common
denominator, X minus two X +2.

Now the left hand side and the
right hand side of the same. The

denominators are already the
same, so the numerators must be

the same. 4X must equal a X +2

plus B. 6  2.

This is the equation that we can
use to find the values for A&B.

Let me write that down again.

We've got 4X is
a X +2.

Plus BX minus
2.

What's a sensible value to put
in for X? Well, if we choose, X

is 2, will lose the second term.

So X is 2 is a good
value to put in here.

Faxes 2 on the left hand side
will have 4 * 2, which is 8.

If X is 2, will have 2 +
2, which is 44A.

And this term will vanish.

So 8 equals 4A.

A will equal 2 and that's the
value for A.

What's another sensible value to
put in for X? Well, if X is

minus two, will have minus 2 +
2, which is zero. Will lose this

term. So if X is
minus two, will have minus 8

on the left.

This first term will vanish.

An ex being minus two here will
have minus 2  2 is minus four

so minus 4B.

So be must also be
equal to two.

So now we've got values for A
and for be both equal to two.

Let's take us back and see what

that means. It means that when

we express. This quantity
4X over X squared minus

four in partial fractions
like this, the values of

A&B are both equal to two.

So the integral were working out
is the integral of X +2 over X

minus 2 + 2 over X +2.

So there we
are. We've now

got three separate
integrals to evaluate,

and it's straightforward
to finish this

off. The integral of X
is X squared divided by two.

The integral of two over X minus
2 equals 2 natural logarithm of

the modulus of X minus 2.

And the integral of two over X
+2 is 2 natural logarithm of X

+2 plus a constant of

integration. If we wanted to do,
we could use the laws of

logarithms to combine these log
rhythmic terms here, but I'll

leave the answer like that.