$\int \frac{\cos 2x}{(\cos x + \sin x)^2} \, dx = $

$\int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx =$

For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,then $y\left(\frac{\pi}{4}\right)$ is equal to:

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

$\int \frac{d x}{x+\sqrt{x-1}} = $

$\int \frac{3^x}{\sqrt{1-9^x}} d x=$