We could write our original integral as being equal to the integral of 1 7th times seven times the square root of seven x plus nine dx. And if we want to, we could take the 1 7th outside of the integral. We don't have to, but we could rewrite this as 1 7th times the integral of seven times the square root of seven x plus nine dx. So now if we set u equal to seven x plus nine, do we have its derivative laying around? Well, sure, the seven is right over here. We know that du, if we want to write in differential form, du is equal to seven times dx. So du is equal to seven times dx.