Answer
$$\int x\,ln^{2}x\,dx=\frac{x^{2}ln^{2}x}{2}-\frac{x^{2}lnx}{2}+\frac{x^{2}}{4}+C$$
Work Step by Step
${(\frac{x^{2}ln^{2}x}{2})}'=x\,ln^{2}x+x\,lnx$
$$\int x\,ln^{2}x\,dx=\frac{x^{2}ln^{2}x}{2}-\int x\,lnx\,dx$$
${(\frac{x^{2}lnx}{2})}'=x\,lnx+\frac{x}{2}$
$$\int x\,ln^{2}x\,dx=\frac{x^{2}ln^{2}x}{2}-[\frac{x^{2}lnx}{2}-\frac{1}{2}\int x\,dx]$$
$$=\frac{x^{2}ln^{2}x}{2}-\frac{x^{2}lnx}{2}+\frac{x^{2}}{4}+C$$
