We want to determine the antiderivative
or evaluate the definite integral
using integration by parts.
Now this problem is actually little bit unique,
because it can be done using some fancy u substitution,
which I believe I did show in a previous video.
But I also want to show how to use integration by parts
to evaluate this indefinite integral.
So here's the integration by parts formula.
And the idea here is to let u equal part of the integral
and then let dv equal the other part.
And then we'll differentiate u to determine
differential u and we'll integrate dv to determine v.
This one's a little tricky because if we want to select u,
so that differential u is simpler than u,
and we also want to select dv
so that it's easy to integrate.
And x as well as the square root of x plus two,
but x and the square root of x plus two
can both be differentiated or integrated.
But before we set this up, let's go ahead and rewrite this
as the integral of x times the quantity x plus two
to the 1/2 power dx.
So notice how if this x wasn't here,
we could actually integrate this using u substitution.
So if we let u equal x, notice that differential u
would just be one dx.
So we'll try that first.
We'll let u equal x, therefore differential u
would just be one dx or dx.
So if this is u,
that means dv must be the quantity x plus two to the 1/2 dx.
So we have to integrate this to determine v.
So let's take a closer look at the integral of
the quantity x plus two to the 1/2 power dx.
It looks like it's going to require u substitution
where u would be equal to x plus two,
but notice how differential u is just one dx or dx.
So we could write this as u to the 1/2
and dx is just equal to du.
So the antiderivative would be u to the 3/2 divided by 3/2
or the quantity x plus two to the 3/2
divided by 3/2 plus c,
which is the same as 2/3 times
the quantity x plus two to the 3/2 plus c.
But, of course, for v, we'll leave the c off
until the very end.
So v is equal to 2/3 times the quantity
x plus two to the 3/2.
And now we can apply the integration by parts formula.
The original integral is equal to u times v,
well here's u and here's v,
so we'll have 2/3 x times the quantity x plus two to the 3/2
minus the integral of v du.
Well here's v and here's du,
so we'll have 2/3 times the integral of
the quantity x plus two to the 3/2 dx.
Notice how this integral here does not have
that extra factor of x that was causing a problem,
so this integral is much easier to evaluate
than the original.
So we'll have 2/3 x times the quantity
x plus two to the 3/2 minus 2/3
times the antiderivative of the quantity
x plus two to the 3/2.
And again, if we let u equal x plus two, du is still dx.
So we can think of this as u to the 3/2 du.
So we'll add one to the exponent.
3/2 plus one would be 5/2
divided by 5/2
which is the same as 2/5 times the quantity
x plus two to the 5/2 power.
And, of course, plus c.
Let's go ahead and clean this up.
We have 2/3 x times the quantity x plus two to the 3/2.
This is going to be minus 4/15
times the quantity x plus two to the 5/2 plus c.
I think this is a fairly challenging one,
because it is fairly difficult to determine
what we should let u equal and what we should let dv equal.
But here is our antiderivative.
In the next video we'll take a look
at applying integration by parts twice
to evaluate an integral.
I hope this was helpful.