Prove

$\tan ^{2}(2 x-3)=\sec ^{2}(2 x-3)-1$

Let $2 x-3=t$

$\therefore 2 d x=d t$

$\Rightarrow \int \tan ^{2}(2 x-3) d x$

$=\int\left[\sec ^{2}(2 x-3)-1\right] d x$

$=\int\left[\sec ^{2} t-1\right] \frac{d t}{2}$

$=\frac{1}{2}\left[\int \sec ^{2} t d t-\int 1 d t\right]$

$=\frac{1}{2}[\tan t-t+\mathrm{C}]$

$=\frac{1}{2}[\tan (2 x-3)-(2 x-3)+\mathrm{C}]$