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\dfrac{1}{x-a} = b+ \dfrac{1}{c+ \dfrac{1}{b+...}}}
\Rightarrow \dfrac{(ba+1)-bx}{x-a}= \dfrac{1}{c+ \dfrac{1}{b+...}}}
\displaystyle\int^{ \frac{\pi}{2}}_0 \dfrac{sinx}{cosx+sinx} \dx
\displaystyle \begin{equation*}\int_a^b f(x) \, \mathrm{d}x = \int_a^b f(a+b-x) \, \mathrm{d}x\end{equation*}