Answer
$$\eqalign{
& {\text{a}}{\text{.}}\frac{1}{2}{x^2}\ln {x^2} - \frac{1}{2}{x^2} + C \cr
& {\text{b}}{\text{.}}\frac{1}{2}{x^2}\ln {x^2} - \frac{1}{2}{x^2} + C \cr
& {\text{c}}{\text{.The results are the same}} \cr} $$
Work Step by Step
$$\eqalign{
& {\text{a}}{\text{. }}\int {x\ln {x^2}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ then}} \cr
& \int {\left( {\ln {x^2}} \right)} xdx = \int {\ln u\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\ln u} du \cr
& {\text{Use the Integral of }}\ln x\left( {{\text{See page 519}}} \right):{\text{ }} \cr
& \int {\ln xdx} = x\ln x - x + C,{\text{ then}} \cr
& \frac{1}{2}\int {\ln u} du = \frac{1}{2}u\ln u - \frac{1}{2}u + C \cr
& {\text{Back - substitute }}u = {x^2} \cr
& \int {x\ln {x^2}} dx = \frac{1}{2}{x^2}\ln {x^2} - \frac{1}{2}{x^2} + C \cr
& \cr
& {\text{b}}{\text{. }}\int {x\ln {x^2}} dx \cr
& {\text{Use integration by parts, }} \cr
& {\text{let }}u = \ln {x^2},{\text{ }}du = \frac{{2x}}{{{x^2}}}dx = \frac{2}{x}dx \cr
& dv = xdx,{\text{ }}v = \frac{1}{2}{x^2} \cr
& {\text{Therefore,}} \cr
& \int {x\ln {x^2}} dx = \left( {\ln {x^2}} \right)\left( {\frac{1}{2}{x^2}} \right) - \int {\frac{1}{2}{x^2}\left( {\frac{2}{x}} \right)dx} \cr
& = \frac{1}{2}{x^2}\ln {x^2} - \int {xdx} \cr
& = \frac{1}{2}{x^2}\ln {x^2} - \frac{1}{2}{x^2} + C \cr
& \cr
& {\text{c}}{\text{. The results are the same}} \cr} $$
