If $$f\!\left( \frac{3x-4}{3x+4} \right) = x + 2,$$​ then $$\int f(x)\,dx$$​ is equal to::

Correct option is C

Given:
​$$f\!\left( \frac{3x-4}{3x+4} \right) = x + 2$$​​
Formula used:
If $$y = \frac{ax+b}{cx+d}$$​, then $$x = \frac{dy-b}{a-cy}$$​​
Solution:
Let $$t = \frac{3x-4}{3x+4}$$​​
Then,
​$$t(3x+4) = 3x-4\\3tx + 4t = 3x - 4\\3x(t-1) = -4(1+t)\\x = \frac{4(1+t)}{3(1-t)}$$​​
So,
​$$f(t) = x + 2\\= \frac{4(1+t)}{3(1-t)} + 2\\= \frac{4(1+t) + 6(1-t)}{3(1-t)}\\= \frac{10 - 2t}{3(1-t)}$$​​
Hence,
​$$f(x) = \frac{10 - 2x}{3(1-x)}$$​​
Now integrate:
​$$\int f(x)\,dx= \int \frac{10 - 2x}{3(1-x)}\,dx\\[2pt]= \frac{1}{3}\int \frac{8 + 2(1-x)}{1-x}\,dx\\[2pt]= \frac{1}{3}\int \left( \frac{8}{1-x} + 2 \right) dx\\[2pt]= \frac{8}{3}\log|1-x| + \frac{2}{3}x + c\\[2pt]= \frac{8}{3}\log|x-1| + \frac{2}{3}x + c\\[2pt]$$​​
The correct answer is (c) $$\frac{8}{3}\log|x-1| + \frac{2}{3}x + c.$$​​