Explanation

(b)
Given, $f\left( \frac{3x - 4}{3x + 4} \right) = x + 2$
Let $\frac{3x - 4}{3x + 4} = X$
$\Rightarrow \ \frac{3x - 4 + 3x + 4}{3x - 4 - (3x + 4)} = \frac{X + 1}{X - 1}$
$\Rightarrow \frac{6x}{- 8} = \frac{X + 1}{X - 1}$
$\Rightarrow x = \frac{4(X + 1)}{3(X - 1)}$
$\therefore\ f(x) = \frac{4(X + 1)}{3(X - 1)} + 2 = \frac{10 - 2X}{3(1 - X)}$
$= \frac{2}{3}\left\lbrack \int_{}^{}{dx} + \int_{}^{}{\frac{4}{1 - x}\ dx} \right\rbrack$
$= \frac{2}{3}\left\lbrack x - 4\log{(1 - x)} \right\rbrack + c$
$= \frac{2}{3}x - \frac{8}{3}\log{(1 - x) + c}$