$\int \frac{\sin x+\cos x}{\sin x-\cos x} d x=$

$\int {\sqrt {{e^x} - 1} } dx = $

Integrate the function: $f^{\prime}(ax+b)[f(ax+b)]^n$

$\begin{aligned} & \int \frac{dx}{(2 \sin x+\sec x)^4}=A(1+\tan x)^{-5} \\ & +B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k, \text{ then } \\ & A+B+C= \end{aligned}$

$\int \frac{x}{\left(1-x^2\right) \sqrt{2-x^2}} d x=$

If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} \,d x=\operatorname{a} \sin^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$,where $c$ is a constant of integration,then the ordered pair $(a, b)$ is equal to