Answer
$\int\frac{dx}{sec(x)-1} = -csc(x)-cot(x)-x+C$
Work Step by Step
$\int\frac{dx}{sec(x)-1} $
$\int(\frac{dx}{sec(x)-1} \times \frac{sec(x)+1}{sec(x)+1}) = \int\frac{sec(x)+1}{sec^2(x)-1}dx$ = $\int\frac{sec(x)}{sec^2(x)-1}dx + \int\frac{1}{sec^2(x)-1}dx$
$sec^2(x) = tan^2(x)+1$
$\int\frac{sec(x)}{tan^2(x)}dx + \int\frac{1}{tan^2(x)}dx = \int cot(x)csc(x)dx + \int cot^2(x)dx = \int cot(x)csc(x)dx + \int [csc^2(x)-1]dx = (-csc(x) + C) + (-cot(x) - x + C) = -csc(x)-cot(x)-x+C$
