Question

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)

• 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
$\frac{x-4}{x+2}=y\Rightarrow \frac{x+2-6}{x+2}=y\Rightarrow 1-\frac{6}{x+2}=y\Rightarrow 1-y=\frac{6}{x+2}\Rightarrow x+2=\frac{6}{1-y}\Rightarrow x=\frac{6}{1-y}-2\Rightarrow 2x+1=-\frac{12}{y-1}-3$
So,
$f\left(y\right)=-\frac{12}{y-1}-3\Rightarrow f\left(x\right)=-\frac{12}{x-1}-3$
Therefore,
$\int f\left(x\right)dx=\int -\frac{12}{x-1}-3dx=-12\ln |x-1|-3x+C$