If, fx−4x+2=2x+1,(x∈R−{1,−2}) then, ∫f(x)dx is equal to : (where C is a constant of integration

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

1. A

$12{\mathrm{log}}_{e}|1-x|-3x+C$

2. B

$-12{\mathrm{log}}_{e}|1-x|-3x+C$

3. C

$-12{\mathrm{log}}_{e}|1-x|+3x+C$

4. D

$12{\mathrm{log}}_{e}|1-x|+3x+C$

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### Solution:

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

Put, $\frac{x-4}{x+2}=t⇒x-4=2t+xt$

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

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

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

On integrating (ii), we get

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

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