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

A
$12 \log_{e} | 1 - x | - 3 x + C$
B
$- 12 \log_{e} | 1 - x | - 3 x + C$
C
$- 12 \log_{e} | 1 - x | + 3 x + C$
D
$12 \log_{e} | 1 - x | + 3 x + C$

Solution:

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

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

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

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

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

On integrating (ii), we get

$\begin{array}{l} \int f (x) d x = - 4 \int \frac{(x + 2)}{x - 1} d x + \int 1 d x = - 4 \int \frac{x + 3 - 1}{x - 1} d x + x \\ = - 4 \int 1 d x - 12 \int \frac{1}{x - 1} d x + x = - 3 x + 12 \int \frac{1}{1 - x} d x \\ = - 3 x + 12 \log_{e} | 1 - x | + C \end{array}$

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

Solution:

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