Integral of Secant function

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$\int \sec\left(9x-9\right)\; dx$

**Answer**

$\int \sec\left(9x-9\right)\; dx$

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

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

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

$=\frac{1}{9}\ln \left|\sec u + \tan u \right| + C$

Substituting back $u=9x-9$

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