Integrate the function $\tan ^{2}(2 x-3)$.

(N/A) We need to evaluate the integral $\int \tan ^{2}(2 x-3) \, dx$.
Using the trigonometric identity $\tan ^{2} \theta = \sec ^{2} \theta - 1$,we can rewrite the integral as:
$\int \tan ^{2}(2 x-3) \, dx = \int (\sec ^{2}(2 x-3) - 1) \, dx$
Now,we split the integral into two parts:
$= \int \sec ^{2}(2 x-3) \, dx - \int 1 \, dx$
Let $t = 2x - 3$. Then $dt = 2 \, dx$,which implies $dx = \frac{1}{2} \, dt$.
Substituting this into the integral:
$= \frac{1}{2} \int \sec ^{2} t \, dt - x + C$
Since $\int \sec ^{2} t \, dt = \tan t + C$,we have:
$= \frac{1}{2} \tan t - x + C$
Substituting back $t = 2x - 3$:
$= \frac{1}{2} \tan (2 x-3) - x + C$
where $C$ is an arbitrary constant.