Let’s have a look at our second example.
We are now dealing with X to the
power of 4 multiplied ln of 3X.
Let’s write this out firstly,
so I have X to the power of 4 and
that is being multiplied by ln of 3X.
Hopefully, I have had a little
thought about substitution.
This won’t really work with
substitution because U would be 3X
and you are not really going to
make your function any easier
when you use substitution.
I am going to find my formula
for integration by parts.
So, uv' equals to uv minus
the integral of vdu dx.
Don’t tend to put the X’s there.
And remember that this is the only part
I’m doing once I’ve split it into uv'.
Now, I’m going to show you a little mistake
that we can make here before we start.
Imagine now if I split this into U and V.
So remember that this isn’t
the correct way of doing it,
I’m just going to show you
something that you could do wrong
and then how to
If I say U equals to x
to the 4 just like we
did previously because
that how it’s written
and V dashed equals to ln of 3x.
This, I can differentiate, no problem, but
do you know the integral of ln of 3x?
We know that ln differentiates to 1 over
X, but we don’t know what it integrates to.
We can divide the integral of
this using integration by parts,
but it’s a much longer expression.
So in this case, I can stop and think,
maybe this isn’t the right way for me,
maybe integrating ln of
x isn’t the right idea.
I can differentiate ln of x.
So in this case, I almost have no choice.
I have to change my options.
So I’m going to make U into
ln of 3X
and vx to the 4.
And that’s only because
I don’t have a choice.
That’s the only way I can do this.
So if I rewrite my V dashed here,
so remember you are splitting
it into U and V dashed.
I have said now that U equals to ln
of 3X and V dashed is X to the fourth.
You may think that this is going
to grow when we integrate this,
the power of X to the fourth will
become bigger, which it will.
But once again, we don’t
have an option here,
so we are just going to go with this
and hope that it becomes easier.
Okay, remember what we do here.
This, I’m going to differentiate
and this, I’m going to integrate.
In order to get from U to U dashed,
I need to differentiate it.
In order to go from V dashed to V, I’m going
to integrate it, so I’m going backwards.
U' here, ln of anything just
goes to 1 over that thing.
So if it’s ln of 3X,
we get 1 over 3X,
but don’t forget this is a
function of a function.
So you have a 3X on the inside, so you
have to multiply it with the 3.
If I tidy this up, I will get --
the 3’s can cancel out, so the
answer to this is just 1 over X,
nice and straightforward.
I’m going to integrate my V dashed.
So when I integrate this, you add 1 to
the power and you divide by new power.
Don’t worry about adding your plus C here.
The C is just a constant, a number, that we
can put right at the end of the integral.
Putting it all into our formula,
so I’m just going to use this
part of the formula here.
So I will say that UV.
So here is my U and here is my V, so those
2 terms multiplied together firstly.
X to the 5 over 5
multiplied by ln of 3x.
I don’t have to do anything
with this anymore,
minus the integral.
And now, my integral of vdu dx
so it’s those 2 terms here.
So I got V multiplied by dU by dX,
so I have X to the 5 over 5
multiplied by 1 over x, dx.
This perhaps to look at.
At first instance isn’t good news
because we still got 2 functions of X’s
here that are multiplying together,
but the nature of these functions
of X’s makes it straightforward.
Because if you think about it,
it’s X to the 5 divided by X.
You can cancel this X to get X to the 4.
So, we can combine this into one function,
which is what we want because those are the
kind of functions that we can integrate.
So if I rewrite this again to
show you what we are left with,
I have the integral of one of this X’s has
canceled with one of the 5 at the top.
So you have X to the 4 over 5 dx.
And again, we have successfully
simplified our integral.
So rather than dealing with
two functions of X here,
X to the 4 times ln of 3x,
we just have one function
to solve to integrate.
I can take this 5 out if
this looks unpleasant.
So, I have ln of 3X.
Because the 5 is just a constant,
I can write it like this and
then I have x to the 4, dx.
Now, I just have to integrate X to the
4, which we know is straightforward,
add 1 to the power
divided by new power.
So, this stays x to the
5 over 5, ln of 3X.
We don’t have to do anything
there, minus 1 over 5, that stays.
When you integrate X to the 4,
that goes to X to the 5 over 5.
So, add 1 to the power
divided by new power.
And lastly, we just simplify that and put
a plus c at the end, x to the 5, ln of 3x
minus this 5s at the
bottom can multiply.
So X to the 5 over 25 and then
you can put a plus C at the end.
If this was a definite
integral and it had limits,
all you do at this stage
once you've integrated it
is just put limits into
this entire integral.
So put X’s everywhere and then
take away the lower limit.
So now, it’s your turn to practice some of
these questions using integration by parts.
You might be tempted to use
substitution on the last question,
but just for the sake of practice, I would
say do them all by integration by parts.
Good luck and I will see you
in the exercise lecture.