00:01
All right, so we're gonna be integrating the following integral 4x cubed over x to a fourth plus one squared now what i'm gonna do right away and you could wait to do this until the next step, but i'm just gonna do it right away we could write this as 4x cubed times x to the fourth plus one all to the negative two with respect to x now the only reason i'm doing this is we're ultimately going to do it.
00:28
So i'm just getting it out of the way early now notice we have something kind of crazy trap by a power, which is why we're going to call you to be x to fourth plus one.
00:41
Now i'm committed to turning this problem into terms of you, meaning i would like a du in place of that.
00:48
The way we get that is by deriving this function right here, u equals x of 4th plus one.
00:56
Now, when i derive, i would get du over dx, but we're going to think about having multiplied up by dx already.
01:03
So what i'll get is du equals 4x cubed dx.
01:08
So like i said, you're kind of thinking about this dx having already migrated up over there via multiplication...