Finding the Integral of sec(x)

So, lets say you want to find the integral of sec(x).

$$\int\mathrm{\sec{(x)}}\, \mathrm{d}x$$ Initially, this seems really hard, and it kinda is. You have to multiple $\sec(x)$ by $\frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)}$. Now, you have: $$\int\mathrm{\sec{(x)}\frac{\sec{(x)} + \tan{(x)}}{\sec{(x)} + \tan{(x)}}}\, \mathrm{d}x$$ $$=\int\mathrm{\frac{\sec^2{(x)} + \sec{(x)}\tan{(x)}}{\sec{(x)} + \tan{(x)}}}\, \mathrm{d}x$$

Note that the numerator is the derviative of $\tan{(x)} + \sec{(x)}$, so you can do u-substitution. $$u = \sec{(x)} + \tan{(x)}, du = (\sec^2{(x)} + \sec{(x)}\tan{(x)})dx$$
$$\int\mathrm{\frac{1}{u}}\, d{u}$$ $$= \ln{|u|} + C$$ $$= \ln{|\sec{(x)} + \tan{(x)}|} + C$$