Question:

<p><strong>Step 1: Find f(x) explicitly</strong></p><p>Given: $f\left(\frac{3x-4}{3x+4}\right) = x + 2$</p><p>Let $t = \frac{3x-4}{3x+4}$. We need to express $x$ in terms of $t$:</p><p>$t(3x+4) = 3x - 4$</p><p>$3tx + 4t = 3x - 4$</p><p>$3tx - 3x = -4 - 4t$</p><p>$x(3t - 3) = -4(1 + t)$</p><p>$x = \frac{-4(1+t)}{3(t-1)} = \frac{-4(1+t)}{3(t-1)}$</p><p></p><p><strong>Step 2: Substitute into the functional equation</strong></p><p>$f(t) = x + 2 = \frac{-4(1+t)}{3(t-1)} + 2$</p><p>$f(t) = \frac{-4(1+t) + 6(t-1)}{3(t-1)}$</p><p>$f(t) = \frac{-4 - 4t + 6t - 6}{3(t-1)}$</p><p>$f(t) = \frac{2t - 10}{3(t-1)} = \frac{2(t-5)}{3(t-1)}$</p><p></p><p><strong>Step 3: Rewrite f(x) for integration</strong></p><p>$f(x) = \frac{2(x-5)}{3(x-1)} = \frac{2(x-1) - 8}{3(x-1)} = \frac{2(x-1)}{3(x-1)} - \frac{8}{3(x-1)}$</p><p>$f(x) = \frac{2}{3} - \frac{8}{3(x-1)}$</p><p></p><p><strong>Step 4: Integrate f(x)</strong></p><p>$\int f(x)\,dx = \int \left(\frac{2}{3} - \frac{8}{3(x-1)}\right)dx$</p><p>$= \frac{2x}{3} - \frac{8}{3}\log|x-1| + C$</p><p>$= -\frac{8}{3}\log|x-1| + \frac{2}{3}x + C$</p><p></p><p><strong>Step 5: Express in terms of $\log|1-x|$</strong></p><p>Note that $\log|x-1| = \log|-(1-x)| = \log|-1| + \log|1-x| = \log|1-x|$</p><p>So: $\int f(x)\,dx = -\frac{8}{3}\log|1-x| - \frac{2}{3}x + C$</p><p></p><p><strong>Step 6: Match with given form</strong></p><p>Given: $\int f(x)\,dx = A\log|1-x| + Bx + C$</p><p>Comparing: $A = -\frac{8}{3}$ and $B = -\frac{2}{3}$</p><p></p><p>$\therefore$ Answer: D</p>