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

Solution:

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

Put, $\frac{x-4}{x+2}=t\Rightarrow x-4=2t+xt$

$\begin{array}{l}\Rightarrow x-xt=2t+4\Rightarrow x(1-t)=2(t+2)\\ \Rightarrow x=\frac{2(t+2)}{1-t}\end{array}$

$\therefore$ (i) becomes $f\left(t\right)=2\left(\frac{2(t+2)}{1-t}\right)+1=\frac{-4(t+2)}{t-1}+1$

or $f\left(x\right)=\frac{-4(x+2)}{x-1}+1$                             …(ii)

On integrating (ii), we get

$\begin{array}{l}\int f\left(x\right)dx=-4\int \frac{(x+2)}{x-1}dx+\int 1dx=-4\int \frac{x+3-1}{x-1}dx+x\\ =-4\int 1dx-12\int \frac{1}{x-1}dx+x=-3x+12\int \frac{1}{1-x}dx\\ =-3x+12{\log }_e|1-x|+C\end{array}$