Question
Integrate the function: tan2(2x - 3)
$ = \int {\left\{ {{{\sec }^2}\left( {2x - 3} \right) - 1} \right\}} dx$
$ = \int {{{\sec }^2}\left( {2x - 3} \right)} dx - \int {1dx} $
Using $\int {{{\sec }^2}\left( {ax + b} \right)} dx = \frac{{\tan \left( {ax + b} \right)}}{a} + c$
$= \frac{{\tan \left( {2x - 3} \right)}}{2} - x + c$
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$\int\frac{\log(\text{x}+2)}{(\text{x}+2)^2}\text{dx}$