Mathematics - Indefinite Integration Question with Solution | TestHub

Suppose, $\frac{x-4}{x+2}=y\Rightarrow x-4=y(x+2)$ $$\Rightarrow x(1-y)=2y+4\Rightarrow x=\frac{2y+4}{1-y}$$ So, $f(y)=2\left(\frac{2y+4}{1-y}\right)+1$ Now, $f(x)=2\left(\frac{2x+4}{1-x}\right)+1=\frac{3x+9}{1-x}$ $$\begin{array}{rl}& =\frac{3(x+3)}{1-x}=\frac{3(x-1+4)}{1-x}=-3+\frac{12}{1-x}\\ & \therefore \int f(x)dx=-12{\log }_e|1-x|-3x+c\end{array}$$