Integral of square of Secant function

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$\int {\sec}^2\left(9x+9\right)\; dx$

**Answer**

$\int {\sec}^2\left(9x+9\right)\; dx$

Substitute $u=9x+9$. So, $du =9\; dx$

Or, $\frac{1}{9}\; du = dx$

And, $\int {\sec}^2\left(9x+9\right)\; dx = \frac{1}{9}\int {\sec u}^2 \;du$

$=\frac{1}{9}\tan u + C$

Substituting back $u=9x+9$

$=\frac{1}{9}\tan\left(9x+9\right)+C$