Answer
$$\left(\frac{2 \ln (2)-(3 / 4)}{2 \ln (2)-1}, \frac{\ln ^{2}(2)-2 \ln (2)+1}{2 \ln (2)-1}\right)$$
Work Step by Step
Given $$f(x)=\ln x, \quad[1,2]$$
Since
\begin{aligned}
M_{y} &=\rho \int_{a}^{b} x f(x) d x \\
&=\int_{1}^{2} x\ln x d x
\end{aligned}
Let
\begin{align*}
u&=\ln x \ \ \ \ \ \ \ \ \ \ \ dv= xdx\\
du&=\frac{1}{x}dx\ \ \ \ \ \ \ \ \ v=\frac{1}{2}x^2
\end{align*}
Then
\begin{aligned}
\int_{1}^{2} x\ln x d x
&=\frac{x^{2} \ln x}{2}\bigg|_{1}^{2}-\int_{1}^{2} \frac{x}{2} d x \\
&=\frac{x^{2} \ln x}{2}-\frac{x^{2}}{4}\bigg|_{1}^{2}\\
&= 2 \ln (2)-\frac{3}{4}
\end{aligned}
and
\begin{aligned}
M_{x} &=\frac{1}{2} \rho \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{1}^{2}\left(\ln x\right)^{2} d x
\end{aligned}
Let \begin{array}{rlrl}
u & =\ln ^{2} x & d v & =1 \\
d u & =\frac{2 \ln x}{x} & & v=x
\end{array}
Then
$$\int \left(\ln x\right)^{2} d x= x \ln ^{2} x-2 \int \ln x \, d x $$
Let \begin{array}{rlrl}
u & =\ln x & d v & =1 \\
d u & =\frac{1}{x} & & v=x
\end{array}
then $\int \ln x=x\ln x-x$ and
$$ \int \left(\ln x\right)^{2} d x=x \ln ^{2} x-2 x \ln x+2 x$$
hence \begin{aligned}
M_{x} &=\frac{1}{2} \int_{1}^{2} \ln ^{2} x d x \\
&=\left.\frac{1}{2}\left(x \ln ^{2} x-2 x \ln x+2 x\right)\right|_{1} ^{2} \\
&=\frac{1}{2}\left(2 \ln ^{2} 2-4 \ln 2+2\right) \\
&=\ln ^{2}(2)-2 \ln (2)+1
\end{aligned}
and
\begin{align*}
M&=\rho A\\
&=\int_{1}^{2} \ln x d x\\
&=\left.(x \ln x-x)\right|_{1} ^{2}\\
&=2 \ln (2)-1
\end{align*}
then
\begin{align*}
\left(\bar{x}, \bar{y}\right)&=\left(\frac{M_{y}}{M},\frac{M_{x}}{M} \right)\\
&=\left(\frac{2 \ln (2)-(3 / 4)}{2 \ln (2)-1}, \frac{\ln ^{2}(2)-2 \ln (2)+1}{2 \ln (2)-1}\right)
\end{align*}
