Answer
$(\frac{40-27\ln3}{24-18\ln3}, \frac{18\ln3-9(\ln3)^{2}-4}{24-18\ln3})$
Work Step by Step
$M_{x}$ = $\frac{1}{2}\int_1^3{[(x-1)^{2}-(\ln{x})^{2}]}dx$ = $(\frac{1}{3}x^{3}-x^{2}-x-x(\ln{x})^{2}+2x\ln{x})|_1^3)$ = $3\ln3-\frac{3}{2}(\ln3)^{2}-\frac{2}{3}$
$M_{y}$ = $\int_1^3{[x(x-1)-(\ln{x})]}dx$ = $(\frac{1}{3}x^{3}-\frac{1}{2}x^{2}\ln{x}-\frac{1}{4}x^{2})|_1^3$ = $\frac{20}{3}-\frac{9}{2}\ln3$
$A$ = $\int_1^3{[(x-1)-(\ln{x})]}dx$ = $(\frac{1}{2}x^{2}-x\ln{x})|_1^3$ = $4-3\ln3$
so the coordinates of the centroid are $(\frac{40-27\ln3}{24-18\ln3}, \frac{18\ln3-9(\ln3)^{2}-4}{24-18\ln3})$
