If f x - 4 x + 2 = 2 x + 1 ,   x ∈ R - 1 , - 2 , then ∫ f x d x is equal to

We have, $f\left(\frac{x-4}{x+2}\right)=2x+1$

Let $\frac{x-4}{x+2}=X\Rightarrow xX+2X=x-4\Rightarrow x\left(X-1\right)=-4-2X\Rightarrow x=\frac{2\left(2+X\right)}{\left(1-X\right)}$

Putting $x=\frac{2\left(2+X\right)}{\left(1-X\right)}$ in the given functional equation, we get 

$\Rightarrow f\left(X\right)=2\times 2\left\{\frac{2+X}{1-X}\right\}+1=\frac{8+4X+1-X}{1-X}=\frac{3X+9}{1-X}=\frac{3\left(X+3\right)}{\left(1-X\right)}$

Replacing $X\to x$, we get

$f\left(x\right)=\frac{3\left(x+3\right)}{\left(1-x\right)}$

$\Rightarrow \int f\left(x\right)dx=3\int \left(\frac{x+3}{1-x}\right)dx=3\int \frac{4-\left(1-x\right)}{1-x}dx=3\left\{\int \frac{4}{1-x}dx-\int dx\right\}$

$=-12\ln \left|1-x\right|-3x+C$, where $C$ is the constant of integration.