So I already told you that we're gonna apply u substitution, but it's interesting to be able to recognize when to use it. And the key giveaway here is, well I have this x squared plus one business to the third power, but then I also have the derivative of x squared plus one, which is two x right over here. So I could do the substitution. I could say u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero, or just two x. I could write that in differential form. A mathematical hand-wavy way of thinking about it is multiplying both sides by dx. And so you get du is equal to two x dx. And so at least this part of the integral I can rewrite.