We want to integrate using integration by parts
but before we do this, I want to talk about
why we'd use integration by parts.
First, the given integral does not fit
any of the basic integration formulas,
so the next step would be to try
integration by substitution.
But notice if we let u equal x
and du would be equal to dx
and we have this extra factor of x squared
so that won't work.
And if we tried letting u equal x squared,
that may be tempting,
but du would be equal to 2x dx,
and we don't have an extra factor of 2x
even though we do have an x in the exponent.
So substitution fails, and so therefore we want to try
integration by parts.
Here's the formula for integration by parts,
and the guidelines are we want to select u
so that differential u is simpler than u;
and we want to select dv so that it's fairly easy to integrate
to determine v.
So the guideline of choosing u
so that differential u is simpler than u
is probably the most significant.
If we let u equal x squared,
then differential u
would be 2x dx,
and 2x is considered simpler than x squared.
Notice if we selected u equals e to the x,
du would be e to the x dx
so it would be the exact same function, not only simpler.
So this leaves e to the x dx for dv.
And now we'll integrate to determine v.
Well the integral of e to the x with respects to x
is just e to the x.
Now we'll apply the integration by parts formula.
The original integral is equal to u times v,
which would be x squared e to the x,
minus the integral of vdu.
Well here's v and here's du,
So we'll have two times the integral of x e to the x dx.
So looking at the integral that we formed,
the integral of x e to the x dx is simpler than
the integral of x squared e to the x dx because,
notice how here we had x squared, and now we have x.
But unfortunately, in order to evaluate this integral,
we have to apply integration by parts again.
But now u is going to be equal to x
so differential u is going to be equal to 1dx or just dx;
so this leaves dv equal to e to the x dx again,
which means v is going to be e to the x.
When we apply the integration by parts formula
to this integral here, we need to be extremely careful.
We're going to have x squared e to the x
minus two, and then from here we're going to apply
the integration by parts formula
so we'll have two times this entire expansion.
So we'll have u times v, which will be x e to the x,
minus the integral of vdu
Well here's v and here's du
so we just have the integral of e to the x dx.
So we have x squared e to the x.
Here we'll have minus two x e to the x,
then we'll have plus two
times integral of e to the x dx.
So then finally we have x squared e to the x
minus two x e to the x
and then the antiderivative of e to the x is just e to the x
so we'll have plus two e to the x
and then finally plus C.
So here we have our antiderivative
which required integration by parts twice.
And that's going to do it for this video.
I hope you found this example helpful.