Let $A=\underset{0}{\overset{\infty }{\int }}\frac{\log x}{1+x^3}dx$. Then the value of $\underset{0}{\overset{\infty }{\int }}\frac{x\log x}{1+x^3}dx$ in terms of A is

$B=\underset{0}{\overset{\infty }{\int }}\frac{x\log x}{1+x^3}dx=\underset{0}{\overset{\infty }{\int }}\frac{(1+x)\log x-\log x}{1+x^3}dx$

$B=\underset{0}{\overset{\infty }{\int }}\frac{\log x}{x^2-x+1}dx-A ..........\left(1\right)$

$Let I=\underset{0}{\overset{\infty }{\int }}\frac{\log x}{x^2-x+1}dx$

$Put x=\frac{1}{t}$

$\Rightarrow I=\underset{\infty }{\overset{0}{\int }}\frac{\log \frac{1}{t}}{\frac{1}{t^2}-\frac{1}{t}+1}(-\frac{dt}{t^2})$

$I=\underset{{\infty }^{ }}{\overset{0}{\int }}\frac{\log t}{t^2-t+1}dt=-\underset{0}{\overset{{\infty }^{ }}{\int }}\frac{\log t}{t^2-t+1}dt=-I$

$2I=0\Rightarrow \boxed{I=0} Sub in \left(1\right)$

$\boxed{B=-A}$