So after all that substitution,
we are now going to move to our
last type of integration method.
This is probably the most complicated
method that we will study in this course
and it’s called
integration by parts.
Now, this is similar or the kind of
functions that you look at are similar
to the functions that you
can use substitution on.
Basically, it’s two functions
But it really is up to you
whatever method you choose now,
whether you use substitution
or integration by parts.
You will see that the questions
that require integration by parts
are usually a bit more complicated than the
once that you can solve by substitution.
It might just be a process of elimination
where you look at a question,
you started off
and realize maybe this will be easier
if I do this with integration by parts.
But when I do the calculations, I usually
leave this method as my last resort.
I try out all the other methods,
see if I can simplify the equation first
to do it easily, just basic integration.
If not, I then move to
integration by substitution
and if that doesn’t work
or give me a nice answer,
I move to integration by parts.
But a lot of the times, you can do integration
by parts questions by substitution
and you can do the substitution
questions by integration by parts
because they have the same format.
They are two functions
Anyway, if I give you the
definition for this,
when you’re looking at integration
by parts, you have two functions.
So, you have a function
of x and a function of x.
But this time we split
it into two parts.
So that’s why we call it
integration by parts.
We split it into a u-term and
we split it into a v-term,
so into dv by dx.
Now this maybe a little
but we will do this as a numerical example
and then you’ll be able to understand.
Once we have done that, we
differentiate one of the terms
and we integrate to the other terms
and we use the other half of this formula
which says UV minus the
integral of vdu dx or vu dx.
Now have a little look at this formula.
Firstly, we do need to know this in
order to apply integration by parts.
But as we do more
you will see that it’s fairly straightforward
as long as you know the formula.
We just have to find the
different parts and put them in.
So let’s look at
our first example.
Now, you may look at this and think, “I’m
going to try and do this by substitution.”
Think about it.
If you take x as your u, this function
will become ue to the u something.
Now, that doesn’t make
this function any easier.
When we use substitution, we’re
usually trying to cancel things out
to make it easier and make it
in terms of one function effects.
So in this case, we'd have
to use integration by parts.
Now, we’ll break it
down into several steps
and then hopefully by the
end of this example,
you will feel more confident applying
integration by parts to these questions,
but just notice that there is no
other way that we can do this.
It’s not straightforward
so you’re not just adding 1 to
the power dividing by a new power.
It is two functions that
are timesing together,
so we need a more
So, let’s try this once again
by integration by parts.
I have my integral x,
e to the power of x, dx.
I’m just going to remind you of
integration by parts here on the side,
so you have the integral of UV dashed
equals to uv minus the integral of vdu dx.
That means that if you have
any integral in this form,
you can break it into u and v ',
so that’s what we are going to do.
We are going to break this
function down into two parts.
One of them, we will call u and the
other part, we will call v dashed.
Now v ' dashed is a differentiated
version of v ' dv/dx.
If I call the first one u equals
to x and the second one e of x,
I now need to do two
different things to it.
Remember what I need for the formula, I
need u, which is u differentiated,
and I also need v by itself.
So, I will have to
change this to u dashed,
which means I have to differentiate it
and in order to go
from v dashed to v,
think about it, what is the
opposite of differentiation?
So if something has been differentiated
to go to the original function,
we are going to have
to integrate it.
So let me just write
that down here.
We are going to differentiate this function
and we are going to integrate this function.
Now, that’s probably the only
complicated step here,
and it often gets
because in this question, you’re differentiating
and integrating at the same time
and then putting it together.
But as long as you’re clear, you
write it out in a methodical way,
it should be fairly
The first step, I’m going
to differentiate my x.
X differentiates -- not integrate
-- differentiates to 1.
So you bring the power down and
you decrease the power by 1.
The e of x, I now need to integrate.
So I’m going to integrate e of x.
Remember, e of x
integrates to just e of x,
so it doesn’t change,
so it’s nice and easy.
Okay, I have everything.
I have u.
I have u dashed.
I have v dashed and I have v.
So I have every component
that I need for this formula.
So let’s put this in, I don’t need to re-write
this part because that’s what I’m doing.
I’m integrating this function.
I’m just going to go straight
into writing the equals part,
so I’m trying to
find the integral.
So in the first part, I do u times v.
So here’s my u and here is my v, so
I’m going to multiply those two terms.
So, u is x and my
V is E to the x,
minus the integral of vdu/dx.
So this is my v function
and this is my du/dx.
So I’m going to multiply those together,
so I have 1 multiplied by e of x dx.
I should have a dx here
at the end and dx there.
Okay, observe why we've done this.
This, we are not having
to do anything with,
so that's done, that's solved.
But look at my integral this time and see
how it’s different to this question here.
So look at my integral here, I am
just integrating e of x dx.
As compared to the question,
which asks me to integrate
x multiplied by e of x.
I have made it a lot
easier for myself.
I made the integral a lot easier
by using integration by parts.
I now have a very straightforward
integral that I can easily do
and I’ve simplified it, and that’s the
idea behind integration by parts.
You have broken into parts,
so that it’s easier rather than more
complicated to solve these equations.
Remember this one thing in mathematics,
when we’re solving equations,
we are always trying to make
solutions easier for ourselves.
If at any point things are getting
more and more complicated
and the equation is growing,
you’ll probably need to go back and
rethink your solution to make it easier.
So I’m just going to rewrite to show you.
I have xe of x minus --
the one, obviously, doesn’t
make any difference.
So I have e of x dx
and all I need to do
is work this out now.
xe of x were not doing anything,
not differentiating or integrating,
so that can stay and all I have
to do now is integrate e of x.
e of x integrates to just e of x
and then you can put a plus C at the end to
say that this is the end of your integral.
So once again, I have made this function
a lot easier by integrating by parts.
We started off with two
We split them into u and v dashed,
and then we put them
into our formula,
hoping that our integral
would become easier.
If you’re getting to a point where
this integral is not becoming easier,
it’s becoming bigger,
you probably need to
go back to this stage
and need to swap your u and v dashed
to see if it gets any easier.
Remember that when you
you’re making the power
smaller, which is good.
We like that.
When you integrate something,
you’re making the power bigger
because you are adding 1 to the power.
So keep that in mind when
you’re doing these questions,
whether you want to simplify
it and make it smaller
or whether you want your function to
become bigger with the powers of x.
There is one last little thing
that I can do here perhaps.
I can take a common factor of e of x,
leaving me with x minus 1
and then, don’t forget your plus C is
just that final constant that we put
at the end of