Answer
$$\sqrt {2\theta - \pi } + \tan \left( {2\theta - \pi } \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\left( {\frac{1}{{\sqrt {2\theta - \pi } }} + 2{{\sec }^2}\left( {2\theta - \pi } \right)} \right)} d\theta \cr
& = \int {\frac{1}{{\sqrt {2\theta - \pi } }}} d\theta + \int {2{{\sec }^2}\left( {2\theta - \pi } \right)} d\theta \cr
& = \int {{{\left( {2\theta - \pi } \right)}^{ - 1/2}}} d\theta + \int {2{{\sec }^2}\left( {2\theta - \pi } \right)} d\theta \cr
& {\text{use the substitution method }} \cr
& u = 2\theta - \pi,\,\,\,\,du = 2d\theta,\,\,\,\,\,d\theta = \frac{{du}}{2} \cr
& \int {{{\left( {2\theta - \pi } \right)}^{ - 1/2}}} d\theta + \int {2{{\sec }^2}\left( {2\theta - \pi } \right)} d\theta \cr
& = \int {{u^{ - 1/2}}} \left( {\frac{{du}}{2}} \right) + \int {2{{\sec }^2}u} d\theta \left( {\frac{{du}}{2}} \right) \cr
& = \frac{1}{2}\int {{u^{ - 1/2}}} du + \int {{{\sec }^2}u} du \cr
& = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + \tan u + C \cr
& = {u^{1/2}} + \tan u + C \cr
& {\text{write in terms of }}\theta;{\text{ replace }}u = 2\theta - \pi \cr
& = \sqrt {2\theta - \pi } + \tan \left( {2\theta - \pi } \right) + C \cr} $$
