`int |x|ln|x|dx` equals `(x ne 0)`A. `(x^(2))/(2)ln|x|-(x^(2))/(4)+C`B. `(1)/(2)x|x|lnx+(1)/(4)x|x|+C`C. `-(x^(2))/(2)ln|x|+(x^(2))/(4)+C`D. `(1)/(2)x|x|ln|x|-(1)/(4)x|x|+C`

`int |x|ln|x|dx` equals `(x ne 0)`A. `(x^(2))/(2)ln|x|-(x^(2))/(4)+C`B

1.	`int \|x\|ln\|x\|dx` equals `(x ne 0)`A. `(x^(2))/(2)ln\|x\|-(x^(2))/(4)+C`B. `(1)/(2)x\|x\|lnx+(1)/(4)x\|x\|+C`C. `-(x^(2))/(2)ln\|x\|+(x^(2))/(4)+C`D. `(1)/(2)x\|x\|ln\|x\|-(1)/(4)x\|x\|+C`
Answer» Correct Answer - D Case I If x gt 0, then `\|x\|=x` On applying integration by parts, we get `therefore" "=intx ln x dx=ln x.(x^(2))/(2)-int(1)/(x).(x^(2))/(2)dx` `=(x^(2))/(2).lnx-(x^(2))/(4)+C` Case II If x lt 0, then `\|x\|=-x` `therefore" "int \|x\|ln \|x\| dx = - int x ln (-x) dx` `-{ln(-x).(x^(2))/(2)-(x^(2))/(4)}+C` `-(x^(2))/(2)ln\|x\|+(x^(2))/(4)+C` Combining both cases, we get `(1)/(2)x\|x\|ln\|x\|-(1)/(4)x\|x\|+C`