1. Understand the greatest integer part

Given $f(x)=\frac{x[x]}{1+x^2}, \qquad x\in(1,3).$

Since $x\in(1,3)$, the value of $[x]$ changes as follows:

• For $x\in(1,2)$, $[x]=1$
• For $x\in[2,3)$, $[x]=2$

So we study $f(x)$ piecewise.

2. Case 1: $1<x<2$

Then $f(x)=\frac{x}{1+x^2}.$

Let $g(x)=\frac{x}{1+x^2}.$

Differentiate: $g'(x)=\frac{(1+x^2)-2x^2}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}.$

For $x\in(1,2)$, we have $x>1$, so $1-x^2<0$. Hence $g'(x)<0$ on $(1,2)$, so $g$ is strictly decreasing.

Now check endpoint behavior: $\lim_{x\to1^+}\frac{x}{1+x^2}=\frac{1}{2},$ $\lim_{x\to2^-}\frac{x}{1+x^2}=\frac{2}{5}.$

Since the interval is open at both ends and the function is decreasing, $f((1,2))=\left(\frac{2}{5},\frac{1}{2}\right).$

3. Case 2: $2\le x<3$

Then $f(x)=\frac{2x}{1+x^2}.$

Let $h(x)=\frac{2x}{1+x^2}.$

Differentiate: $h'(x)=\frac{2(1+x^2)-4x^2}{(1+x^2)^2}=\frac{2-2x^2}{(1+x^2)^2}=\frac{2(1-x^2)}{(1+x^2)^2}.$

For $x\in[2,3)$, again $x>1$, so $1-x^2<0$. Hence $h'(x)<0$, so $h$ is strictly decreasing on $[2,3)$.

Now evaluate endpoint behavior: $h(2)=\frac{4}{5},$ $\lim_{x\to3^-}\frac{2x}{1+x^2}=\frac{6}{10}=\frac{3}{5}.$

Since $x=2$ is included, $\frac45$ is included. Since $x=3$ is not included, $\frac35$ is not included. Thus, $f([2,3))=\left(\frac{3}{5},\frac{4}{5}\right].$

4. Combine the two parts

Therefore the total range is $\left(\frac{2}{5},\frac{1}{2}\right)\cup\left(\frac{3}{5},\frac{4}{5}\right].$

Since $\frac{1}{2}<\frac{3}{5},$ these intervals are disjoint.

5. Match with the options

Option A is $\left(\frac{2}{5},\frac{1}{2}\right)\cup\left(\frac{3}{4},\frac{4}{5}\right].$ This is incorrect because the second interval should start at $\frac35$, not $\frac34$.

Option B is $\left(\frac35,\frac45\right),$ which misses the first interval and excludes $\frac45$.

Option C is $\left(\frac25,\frac45\right],$ which incorrectly includes values between $\frac12$ and $\frac35$.

Option D is $\left(\frac25,\frac35\right]\cup\left(\frac34,\frac45\right),$ which is also incorrect.

So the correct range is $\boxed{\left(\frac25,\frac12\right)\cup\left(\frac35,\frac45\right]}.$