If $$f\left(\frac{x-4}{x+2}\right) = 2x + 1$$, $$(x \in R - \{1, -2\})$$, then $$\int f(x)dx$$ is equal to (where C is a constant of integration):

We are told that the function satisfies the relation $$f\!\left(\dfrac{x-4}{x+2}\right)=2x+1$$ for every real $$x$$ except $$x=1,-2$$. Our first aim is to rewrite this statement so that the argument of $$f$$ becomes an independent variable.

Let us set $$t=\dfrac{x-4}{x+2}\,.$$ We now express $$x$$ in terms of $$t$$.

Multiplying both sides by $$(x+2)$$ gives $$t(x+2)=x-4.$$ Expanding and bringing like terms together, we have

$$tx+2t=x-4.$$ Subtract $$x$$ from both sides:

$$tx-x=-4-2t.$$ Factor $$x$$ from the left:

$$x(t-1)=-(4+2t).$$ So, dividing by $$(t-1)$$,

$$x=\dfrac{-(4+2t)}{t-1}.$$

We now substitute this value of $$x$$ into the given formula $$2x+1$$ in order to obtain $$f(t)$$:

$$\begin{aligned} f(t)&=2x+1\\[4pt] &=2\!\left(\dfrac{-(4+2t)}{t-1}\right)+1\\[4pt] &=\dfrac{-(8+4t)}{t-1}+1\\[6pt] &=\dfrac{-(8+4t)}{t-1}+\dfrac{t-1}{t-1}\\[6pt] &=\dfrac{-(8+4t)+t-1}{t-1}\\[6pt] &=\dfrac{-9-3t}{t-1}\\[6pt] &=-3\;\dfrac{3+t}{t-1}. \end{aligned}$$

Finally, replacing the dummy variable $$t$$ by $$x$$ (because a function’s name of the variable is immaterial), we get the explicit formula

$$f(x)=-3\,\dfrac{x+3}{x-1}.$$

Our task is to integrate this function. We rewrite the integrand by performing polynomial division on the fraction $$\dfrac{x+3}{x-1}$$:

Since $$x+3=(x-1)+4$$, we have $$\dfrac{x+3}{x-1}=1+\dfrac{4}{x-1}.$$ Therefore,

$$f(x)=-3\!\left(1+\dfrac{4}{x-1}\right)=-3-\dfrac{12}{\,x-1}.$$

Now we integrate term by term:

$$\int f(x)\,dx=\int\!\left(-3-\dfrac{12}{x-1}\right)dx.$$

We recall the standard integral formula $$\int \dfrac{1}{x-a}\,dx=\ln|x-a|+C.$$ Using this, we get

$$\begin{aligned} \int f(x)\,dx&=\int -3\,dx+\int -\dfrac{12}{x-1}\,dx\\[6pt] &=-3x-12\ln|x-1|+C, \end{aligned}$$

where $$C$$ is the constant of integration. Since $$|x-1|=|1-x|$$, we can equally well write the logarithmic term as $$\ln|1-x|$$. Thus, another perfectly equivalent final form is

$$\boxed{-12\ln|1-x|-3x+C}.$$

Comparing with the given options, this matches Option B. Hence, the correct answer is Option 2.