To treat the integral 1- -44(2x+1)-2 dx as a constant multiple of one of the form unu' dx, we let n = -2, u = 2x + 1, and u' = 2. So we need the factor 2, rather than -44, in the integrand. If we first reorder the factors and then multiply by the constant factor 2 (and divide it out), we have the following. -44(2x+1)-2 dx = (2x+1)-2(-44) dx = (2x + 1)-2(-24)(1 = (2x+1)-2.2 dx The result is a constant multiple of unu' dx. dx