Given,

f$$\left( {{{3x - 4} \over {3x + 4}}} \right)$$ = x + 2,    x  $$ \ne $$ $$-$$ $${4 \over 3}$$

Let, $${{3x - 4} \over {3x + 4}}$$ = t

$$ \Rightarrow $$$$\,\,\,$$ 3x $$-$$ 4 = 3tx + 4t

$$ \Rightarrow $$$$\,\,\,$$ 3x $$-$$ 3tx = 4t + 4

$$ \Rightarrow $$$$\,\,\,$$ x = $${{4t + 4} \over {3 - 3t}}$$

So, f(t) = $${{4t + 4} \over {3 - 3t}}$$ + 2 = $${{10 - 2t} \over {3 - 3t}}$$

$$\therefore\,\,\,$$ f (x) = $${{10 - 2x} \over {3 - 3x}}$$

$$\therefore\,\,\,$$ $$\int {f(x)\,dx} $$

= $$\int {{{2x - 10} \over {3x - 3}}} \,dx$$

= $$\int {{{2x} \over {3x - 3}} - 10\int {{{dx} \over {3x - 3}}} } $$

= $${2 \over 3}\int {{{x - 1} \over {x + 1}}} dx + {2 \over 3}\int {{{dx} \over {x - 1}} - {{10} \over 3}} \int {{{dx} \over {x - 1}}} $$

= $${2 \over 3}$$ x + $${2 \over 3}$$ log $$\left| {x - 1} \right|$$ $$-$$ $${{10} \over 3}$$ log $$\left| {x - 1} \right|$$ + C

= $${2 \over 3}$$ x $$-$$ $${8 \over 3}$$ log $$\left| {x - 1} \right|$$ + C

$$\therefore\,\,\,$$ A = $$-$$ $${8 \over 3}$$ and B = $${2 \over 3}$$