I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.
How did we go from: $\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$ to $\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$
Thank you for your time! $ \int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$ To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$. Integrating by parts, we have: $\begin{align*} \int\csc^2 x \csc x \,dx &= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\ &= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\ &= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &\text{(since }\cot^2 x = \csc^2-1\text{)}\\ &= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\ &= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx \end{align*}$ From $\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx$ we obtain $\begin{align*} \int\csc^3x\,dx + \int\csc^3 x\,dx &= -\cot x \csc x + \int\csc x\,dx\\ 2\int\csc^3 x\,dx &= -\cot x\csc x + \int\csc x\,dx\\ \int\csc^3x\,dx &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\ &=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\ &= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\ &= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\ &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C \end{align*}$