If I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one. So we should make that substitution. If we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x, is equal to two x plus one. If we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand-wavy mathematics, but it's appropriate here. So we could say two x plus one times dx. And now what's really interesting is here is we have our u right over there, and notice we have our two x plus one times dx.