Question
$\int\tan^2(2\text{x}-3)\text{dx}$

Answer

$\int\tan^2(2\text{x}-3)\text{dx}$
$=\int[\sec^2(2\text{x}-3)-1]\text{dx}$
$=\int\sec^2(2\text{x}-3)\text{dx}-\int1\text{dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$

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