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

$-\frac{8}{3} \log _e|1-x|+\frac{2}{3} x+C$

We have, $f\left(\frac{3 x-4}{3 x+4}\right)=x+2$

Let $\frac{3 x-4}{3 x+4}=\alpha$

$\Rightarrow \frac{(3 x-4)+(3 x+4)}{(3 x-4)-(3 x+4)}=\frac{\alpha+1}{\alpha-1}$

$\Rightarrow \frac{6 x}{-8}=\frac{\alpha+1}{\alpha-1}$

$\Rightarrow x=-\frac{4}{3}\left(\frac{\alpha+1}{\alpha-1}\right)$

$\Rightarrow x+2=-\frac{4 \alpha+4}{3 \alpha-3}+2=\frac{-4 \alpha-4+6 \alpha-6}{3 \alpha-3}=\frac{2 \alpha-10}{3 \alpha-3}$

∴  $f\left(\frac{3 x-4}{3 x+4}\right)=x+2$

$\Rightarrow f(\alpha)=\frac{2 \alpha-10}{3 \alpha-3}$

$\Rightarrow f(\alpha)=\frac{2}{3}\left(\frac{\alpha-5}{\alpha-1}\right)$

$\Rightarrow f(\alpha)=\frac{2}{3}\left(\frac{\alpha-1-4}{\alpha-1}\right)=\frac{2}{3}\left(1-\frac{4}{\alpha-1}\right)=\frac{2}{3}-\frac{8}{3(\alpha-1)}$

$\Rightarrow f(x)=\frac{2}{3}-\frac{8}{3(x-1)}$

∴  $\int f(x) d x=\int\left\{\frac{2}{3}-\frac{8}{3(x-1)}\right\} d x$

$\Rightarrow \int f(x) d x=\frac{2}{3} x-\frac{8}{3} \log _e|x-1|+C$