### Video Transcript

Determine the indefinite integral
of two π₯ squared times by π to the π₯ plus two with respect to π₯.

First, we notice that we have a
factor of two inside our integral. So we can factor this out of the
integral. This leaves us with two times the
indefinite integral of π₯ squared times by π to the π₯ plus two with respect to
π₯. In order to integrate this
integral, we need to use integration by parts. When using integration by parts,
weβre using the formula which tells us that the integral of π’ times dπ£ dπ₯ with
respect to π₯ is equal to π’ times π£ minus the integral of dπ’ dπ₯ times π£ with
respect to π₯.

From our integral, we need to
choose a π’ and a dπ£ dπ₯. When picking our π’ here, we want
to pick a term which would decrease in magnitude when differentiated. And when picking our dπ£ by dπ₯, we
want to pick a term which would decrease or stay the same when integrated. And so therefore, we will pick π’
equals π₯ squared. And dπ£ by dπ₯ is equal to π to
the π₯ plus two.

So now, we can find dπ’ by dπ₯ by
differentiating π’ with respect to π₯. And so thatβs the same as
differentiating π₯ squared with respect to π₯. So this gives us two π₯. And then, in order to find π£, we
simply integrate dπ£ by dπ₯. And so thatβs the same as
integrating π to the π₯ plus two with respect to π₯. And since this is an exponential
term, this is simply π to the π₯ plus two. So now, we have our integral in the
form of the integral of π’ times dπ£ by dπ₯ dπ₯ since π₯ squared is π’ and π to the
π₯ plus two is dπ£ by dπ₯.

Now, weβre ready to use integration
by parts. Using the formula and remembering
our constant term, multiplying the integral, we get that this is equal to two times
by π₯ squared π to the π₯ plus two minus the integral of two π₯ π to the π₯ plus
two with respect to π₯. Now, we notice that we have another
factor of two inside our integral, which can be moved and factored out of the
integral. Next, we look at our integral
again. And we notice that we canβt
integrate this directly. So we will need to use integration
by parts again.

So we can choose π’ and dπ£ by dπ₯
similarly to before, giving us π’ equals π₯. And dπ£ by dπ₯ is equal to π₯ plus
two. Now, we can find dπ’ by dπ₯ and π£
by differentiating π’ and integrating dπ£ by dπ₯. This leaves us with dπ’ by dπ₯ is
equal to one. And π£ is equal to π to the π₯
plus two.

And now, weβre ready to use
integration by parts again, remembering that our π₯ term is π’ and our π to the π₯
plus two term is dπ£ by dπ₯. This leaves us with two times by π₯
squared π to the π₯ plus two minus two times by π₯ π to the π₯ plus two minus the
integral of π to the π₯ plus two with respect to π₯. And now, we have the integral of π
to the π₯ plus two with respect to π₯, which we know how to integrate.

Our next step here is to perform
this integration. Since weβre integrating an
exponential term here, it is just equal to itself. And so this leaves us with two
times by π₯ squared π to the π₯ plus two minus two π₯ times by π to the π₯ plus
two minus π to the π₯ plus two. And now, since this was an
indefinite integral, we mustnβt forget to add on our constant of integration π. Finally, we notice that we have an
π to the π₯ plus two, multiplying each term. And so we can factor this out
here. Now, all that remains to do is to
expand this minus two times by π₯ minus one bracket.

So we obtain a final answer that
the indefinite integral of two π₯ squared times π to the π₯ plus two with respect
to π₯ is equal to two times by π to the π₯ plus two times by π₯ squared minus two
π₯ plus two plus π.