00:01
And our last example which we are also
going to do by integration by parts,
but remember we are going to
have to tweak it a little bit
because you have a
function as a denominator.
00:09
And when we have those types of scenarios,
we can bring it up and see
what we can do with it.
00:14
So my integral here is x over 5x
plus 1 cube, dx.
00:23
Integration by parts
formula states this,
uv' dx equals to uv minus
the integral of vu' dx.
00:40
So, we need every component.
00:41
We need u, v, u' and v'.
00:45
But before I do this, I’m
just going to tidy this is up
and make it more
understandable for myself.
00:49
So, I’m going to bring that up,
5x plus 1 to the minus 3, dx.
00:54
And now, I can clearly
see that it’s a product.
00:56
So you have one function
multiplying by the other function.
01:00
If I say that u equals
to x and v' equals to 5x
plus 1 to the minus 3.
01:08
This needs to differentiate.
01:10
This needs to integrate.
01:12
U' goes to 1 and V, now, be
careful, we are integrating.
01:17
When you integrate,
add 1 to the power divided by new
power divided by the differential.
01:22
So we have 5x plus 1.
01:24
Add 1, so, minus 3 plus
1 gives you minus 2,
divided by new power and divided by the
differential of the inside which is 5.
01:34
So, we get 5x plus 1 to the
minus 2 over minus 10.
01:40
Don’t take this down yet.
01:41
So, don’t take this power down as yet
because we are still going to need to
do some integration, some simplifying,
so just leave it as it is and we
will do that right at the end.
01:51
So put it all into our
formula, we have u times v,
so we have that simplified version here.
01:58
So I have X multiplied by 5x plus
1 to the minus 2 over minus 10,
minus the integral of vdu dx.
02:08
So those two terms multiplied together.
02:11
So that gives me 1 times 5x plus
1 to the minus 2 over minus 10.
02:18
So in some sense, it’s good.
02:19
We didn’t take this down because
we are going to need it
up as a numerator
when we integrate.
02:27
This, I can tidy
up, this term here
because we are not really doing
anything more with this.
02:32
So I can write this as minus X over 10.
02:34
The 10 can move anywhere
and I can also move that down, so I
can write it as 5x plus 1 squared.
02:41
I will tidy that up in the next step.
02:43
We now need to integrate
5x plus 1 to the minus 2 dx,
and the minus 10, I can
take outside as 10.
02:54
Because it was a minus, it
makes this sign a plus,
so it’s minus times a minus there,
which will make it into a plus.
03:01
So let’s integrate,
this can stay as minus x over 10,
5X plus 1 to the power
2 plus 1 over 10.
03:12
Integrate this now, remember
what happens when you integrate,
add 1 to the power divided by new
power divided by the differential.
03:20
So 5x plus 1 to the minus 1,
so minus 2 plus 1 is minus 1
divided by minus 1 times by the differential
of the inside which is 5x plus 1.
03:32
I’ll put my plus C at the end once
I have tidied this up a little bit.
03:35
So I've got minus X over 10,
5x plus 1 squared and
then this minus here
will make this entire term minus.
03:45
So imagine this minus is multiplying with
this 10 which is then multiplying here.
03:50
So this will make this
entire term minus.
03:54
The 10 and the 5 at the bottom can
multiply to give you 1 over 50
and then I have 5x
plus 1 to the minus 1 which I can bring
down as 5x plus 1 to the positive 1
and I will put plus C at the end.
04:09
So we have integrated two functions
that are dividing each other
using integration by parts
just by bringing it up.
04:17
A few last tips, just remember that try
or think about substitution before.
04:23
If you think substitution will make
it a little bit more complicated,
then move to
integration by parts.
04:29
Always look at your u' and v', so your
first choices here are fairly important.
04:35
It is common to get
them wrong sometimes,
but the only thing that you need to be
careful about is when you have ln's.
04:42
Because an ln function should
never be your v' function,
it should always be u, because we
don’t know the integral of ln of x.
04:50
You can do it.
04:51
You can do it using
integration by parts,
but it gives a much longer
and complicated result.
04:57
So only when you have a function and ln,
in that case make sure
that ln becomes your u
and everything else
becomes your v'.