Answer
$$\ln \left| {1 - \cot \theta } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta \left( {\tan \theta - 1} \right)}}} d\theta \cr
& {\text{Let }}x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta \left( {\tan \theta - 1} \right)}}} d\theta = \int {\frac{{dx}}{{x\left( {x - 1} \right)}}} \cr
& {\text{Decompose }}\frac{1}{{x\left( {x - 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{x\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{x - 1}} \cr
& 1 = A\left( {x - 1} \right) + Bx \cr
& x = 0 \to 1 = A\left( {0 - 1} \right) \to A = - 1 \cr
& x = 1 \to 1 = B\left( 1 \right) \to B = 1 \cr
& {\text{Then,}} \cr
& \frac{1}{{x\left( {x - 1} \right)}} = \frac{{ - 1}}{x} + \frac{1}{{x - 1}} \cr
& \int {\frac{{dx}}{{x\left( {x - 1} \right)}}} = \int {\left( {\frac{{ - 1}}{x} + \frac{1}{{x - 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }} - \ln \left| x \right| + \ln \left| {x - 1} \right| + C \cr
& = \ln \left| {\frac{{x - 1}}{x}} \right| + C \cr
& {\text{Write in terms of }}\theta \cr
& = \ln \left| {\frac{{\tan \theta - 1}}{{\tan \theta }}} \right| + C \cr
& = \ln \left| {1 - \cot \theta } \right| + C \cr} $$
