00:01
Okay, we want to evaluate this integral of the function that is x cubed over 4 plus x squared to 3 halves.
00:10
And to solve this integral, to evaluate this integral, we will be using u substitution in this case because it gives us the most clear path to get the answer.
00:22
So first thing we're going to do is set ru, and we're going to say that ru is going to be equal to what's inside our parentheses here in the denominator, which is 4 plus x squared.
00:38
So with u equal to 4 plus x squared, this means that du is equal to 2x.
00:45
And we're letting you be this 4 plus x squared, because we're going to be able to substitute in our du in a way that gets rid of this x cubed in the numerator.
00:55
It allows us to break up the integral into something simpler and something we can work with.
01:01
So this means d u is equal to 2x.
01:04
Dx, which in turn means that dx must be d u over 2x, and this allows us to do the substitution and shift from dx to du.
01:20
So this means we can rewrite our integral with what we have.
01:23
So we're going to have a factor of one -half because of what's the denominator of our substitution.
01:27
So we have a factor of one -half out front because it's a constant.
01:31
Integral, still all the x cubed over.
01:37
Now this is u, because again, this is u.
01:40
Substitution, u to three halves d u, we still have a factor of x.
01:47
The factor of x comes again from this substitution we made.
01:51
So we see we have an x cubed in numerator, an x in the denominator.
01:55
Those can cancel with each other and leaves us with an x squared in the numerator.
02:03
And from our u substitution, we can then write that an x squared is the same thing as u minus four.
02:11
So we do another substitution so that we have our integral, our new integral, of purely u variables, which allows us to complete the integral in easier way.
02:23
So we'll write this in red.
02:24
So we still have the one -half.
02:27
The integral of x squared now becomes the u minus four.
02:30
So u minus four in the numerator over u to the three halves, d ,u.
02:39
And now since we have a fraction, a fraction for our integral now, we can break it up and just do term -by -term integration.
02:48
So we end up going to break this up into two separate integrals.
02:51
We'll have u over u to three halves, then minus the integral of four over u to the three halves.
02:58
So this means we still have this factor of one -half.
03:01
That's going to be distributed to both integrals that we're going to end up doing...