1. We are given $f(x)+f\left(\frac{1}{1-x}\right)=1+x, \qquad x\in \mathbb R\setminus\{0,1\}.$

We need to find $f(2)$.

2. Define the transformation $T(x)=\frac{1}{1-x}.$ Then the given relation becomes f(x)+f(T(x))=1+x. \tag{1}

Now compute repeated applications of $T$:

$T(x)=\frac{1}{1-x},$ $T^2(x)=T(T(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{1}{\frac{-x}{1-x}}=\frac{x-1}{x},$ $T^3(x)=T\left(\frac{x-1}{x}\right)=\frac{1}{1-\frac{x-1}{x}}=\frac{1}{\frac{1}{x}}=x.$

So the three values cycle as $x \to \frac{1}{1-x} \to \frac{x-1}{x} \to x.$

3. Write the functional equation for these three inputs.

From (1):

For $x$: f(x)+f\left(\frac{1}{1-x}\right)=1+x. \tag{2}

For $T(x)=\frac{1}{1-x}$: f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=1+\frac{1}{1-x}. \tag{3}

For $T^2(x)=\frac{x-1}{x}$: f\left(\frac{x-1}{x}\right)+f(x)=1+\frac{x-1}{x}. \tag{4}

4. Now substitute $x=2$.

The cycle is: $2 \to \frac{1}{1-2}=-1 \to \frac{2-1}{2}=\frac12 \to 2.$

So equations (2), (3), (4) become:

f(2)+f(-1)=3, \tag{5} $f(-1)+f\left(\frac12\right)=1+(-1)^{-?}$ Better to directly use the original formula carefully:

For $x=-1$, $f(-1)+f\left(\frac{1}{1-(-1)}\right)=1+(-1),$ f(-1)+f\left(\frac12\right)=0. \tag{6}

For $x=\frac12$, $f\left(\frac12\right)+f\left(\frac{1}{1-\frac12}\right)=1+\frac12,$ f\left(\frac12\right)+f(2)=\frac32. \tag{7}

5. Let $a=f(2),\quad b=f(-1),\quad c=f\left(\frac12\right).$ Then from (5), (6), (7):

a+b=3, \tag{8} b+c=0, \tag{9} c+a=\frac32. \tag{10}

Add (8) and (10): $2a+b+c=\frac92.$ Using (9), $b+c=0$, so $2a=\frac92,$ $a=\frac94.$

Hence, $f(2)=\frac94.$

6. Checking options:

• A: $\frac94$ ✅
• B: $\frac74$ ❌
• C: $\frac73$ ❌
• D: $\frac92$ ❌

Therefore the correct answer is A.