Question

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

• A. $12{\log }_e|1-x|-3x+C$
• B. $-12{\log }_e|1-x|-3x+C$
• C. $-12{\log }_e|1-x|+3x+C$
• D. $12{\log }_e|1-x|+3x+C$

Answer 1

The correct option is B $-12{\log }_e|1-x|-3x+C$

Let $\frac{x-4}{x+2}=y\Rightarrow x-4=yx+2y\Rightarrow x(1-y)=2y+4\Rightarrow x=\frac{2y+4}{1-y}$

This gives us $f\left(y\right)=2\left(\frac{2y+4}{1-y}\right)+1$

So, we have $f\left(x\right)=2\left(\frac{2x+4}{1-x}\right)+1=\frac{3x+9}{1-x}=-3\left(\frac{x-1+4}{x-1}\right)=-3-\frac{12}{x-1}$

Thus $\int f\left(x\right)dx=-12{\log }_e|1-x|-3x+c$

So, the correct answer is option B.