Integration by Parts Example 7

Uploaded by TheIntegralCALC on 19.07.2010

Hi everyone! Welcome back to We're going to be doing another integration
by parts problem today. This one is the integral of x times e to the x dx.
And the reason that we know that this is an integration by parts problem is because we
have two components here in the integral that comprise the entire integral. It's not a partial
fractions problem because we don't have a fraction here. It could be u-substitution
but you're always looking for something to cancel when you use u-substitution and you're
always looking for the derivative of u to be something that will cancel out something
that's already in your integral. So for example, if we assign u to e to the x, the derivative
is e to the x. That's not going to cancel out the x here. And if we assign u to x, the
derivative of x is just one and that's certainly not going to cancel anything out.
So let's try integration by parts. Remember the integration by parts equation is the integral
of u times dv equals u v minus the integral of v du.
So this is the equation we're working with and we need to assign u and dv to identities
in our integral here. So what we, the easiest way to do it is to assign u to something that
will become simpler when you take its derivative. So in this case, that's definitely x. The
derivative of e to the x is still e to the x which does not simplify at all but the derivative
of x is one, which is much simpler than x. So we're going to assign u to x, x and then
we need to assign dv, by default to the other value in our problem, e to the x.  And from
u and dv, we need to find du and v. We'll find du by taking the integral, the derivative
of u and the derivative of x is just one. v, we determine by taking the integral of
e to the x. The integral of e to the x is still e to the x.
So now that we've found these four identities, we can go ahead and plug them into the second
part or the right side of this equation here. So you can see, we have u times v. So here,
u times v. So, x times e to the x minus the integral of v, which we have is e to the x
times du, so times one and then we put in the dx here as notation. This one's going
to go away, (right?) because it's just redundant. So we're just left with the integral of e
to the x, which is e to the x. So x times e to the x minus, we take the integral and 
we just flip e to the x and then we add c to account for the constant. And this is actually
our final answer. If you would like to, you can factor out an
e to the x, so you either have x e to the x minus e to the x plus c or if you want to
factor an e to the x, you can do so. I'm not sure which way is simpler.  You'd have x
minus one plus c. Either way, they're equally correct but that's
your final answer. Hope that helped. See you guys next time! Bye!