x2ln3x=∫x31−2ln3xdx, so 2∫x3ln3xdx=∫x31dx−x2ln3x oe. M1 for using integration as reverse of differentiation from part (a); allow their A,B, no extra terms.
∫x31dx=−2x21 oe, nfww. B1, FT on their A.
−4x21−2x2ln3x+c oe. B2; M1 dep for rearranging simplification and attempting to integrate ax−3. A1 must include +c.