Evaluate $\displaystyle\int \dfrac{\cos x}{\cos x - 1} dx$
$\begin{aligned} \text{Let } I & = \int \dfrac{\cos x}{\cos x - 1} dx \\\\ & = \int \dfrac{\cos x - 1 + 1}{\cos x - 1} dx \\\\ & = \int \dfrac{\cos x - 1}{\cos x - 1} dx + \int \dfrac{1}{\cos x - 1} dx \\\\ & = \int dx + \int \dfrac{\cos x + 1}{\left(\cos x + 1\right) \left(\cos x - 1\right)} dx \\\\ & = \int dx + \int \dfrac{\cos x + 1}{\cos^2 x - 1} dx \\\\ & = \int dx + \int \dfrac{\cos x + 1}{- \sin^2 x} dx \\\\ & = \int dx - \int \dfrac{\cos x}{\sin^2 x} dx - \int \dfrac{1}{\sin^2 x} dx \\\\ & = \int dx - \int \cot x \;\; \text{cosec }x \;\; dx - \int \text{cosec}^2 x \; dx \\\\ & = x + \text{cosec }x + \cot x + c \end{aligned}$