Let, $\frac{{2x - 1}}{{x - 3}} = t$
$\therefore x = \frac{{3t - 1}}{{t - 2}}$
$\therefore f(t) = {\left( {\frac{{3t - 1}}{{t - 2}}} \right)^2}$
Now, $\int {f(t)dt} = \int {{{\left( {\frac{{3t - 1}}{{t - 2}}} \right)}^2}dt} $
$ = \int {{{\left[ {\frac{{3(t - 2) + 5}}{{t - 2}}} \right]}^2}dt} $
$ = \int {{{\left( {3 + \frac{5}{{t - 2}}} \right)}^2}dt} $
$ = \int {9 + \frac{{30}}{{t - 2}} + \frac{{25}}{{{{(t - 2)}^2}}}dt} $
$ = 9t + 30\ln |t - 2| - \frac{{25}}{{t - 2}} + C$
The integral asked $=9x + 30\ln |x - 2| - \frac{{25}}{{x - 2}} + C$
