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.