-
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 55-B.
-
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.