Answer Happy

I need help with the highlighted step in the attached question.Where did they get the factor of two that they pulled out?
Problem 3) Evaluate the integral using integration by parts. Solution: Integration by parts splits the integrand into two parts, one part called u which is "destructible" when differentiating and the other part called dv which is the rest of the integrand. Then the rule for this type of integration is: fud 3xlnx dx udv = uv - In this problem, let u = 3x and dv = lnxdx. From the video, you learned that flnx dx = xlnx - x + C.Then, du = 3dx and v= xlnx - x. Integrate the given integral: 3xlnx dx = 3x(xlnx - x) - = 3x(xlnx - x) - -S = 3x(xlnx - x) - 3xlnx dx = 3 Answer: x²Inx - ³x² + C vdu f(3xl (3xlnx - 3x) dx 3xlnx dx + S 3x dx 3xlnx dx + x² +37x² 2 3 3x(xlnx − x) + x² + C 1 3 3xlnx dx = x = (3x²lnx − 3x²) + ²√x² + C 3 3 = 2x²1x − 3x² + ³x² + C 2 4 3 3 =x²Inx-²x² + C How?