$\begin{array}{l}
\frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, x \neq 0,1 \\
\Rightarrow \frac{x^{2}+(x-1)^{2}}{x(x-1)}=\frac{17}{4} \\
\Rightarrow \frac{x^{2}+x^{2}-2 x+1}{x^{3}-x}=\frac{17}{4} \\
\Rightarrow \frac{2 x^{2}-2 x+1}{x^{2}-1}=\frac{17}{4} \\
\Rightarrow 8 x^{2}-8 x+4=17 x^{2}-17 x \\
\Rightarrow 9 x^{2}-9 x-4=0 \\
\Rightarrow 9 x^{2}-12 x+3 x-4=0 \\
\Rightarrow 3 x(3 x-4)+1(3 x-4)=0 \\
\Rightarrow(3 x-4)(3 x+1)=0 \\
\Rightarrow 3 x-4=0 \text { or } 3 x+1=0 \\
\Rightarrow x=\frac{4}{3} \text { or } x=-\frac{1}{3}
\end{array}$

Hence, $\frac{4}{3}$ and $-\frac{1}{3}$ are the roots of the given equation.