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

$\int [1+2 \tan x(\tan x+\sec x)]^{\frac{1}{2}} dx = $

Find the following integral: $\int (\sin x + \cos x) \, dx$

If $f\left( \frac{x - 4}{x + 2} \right) = 2x + 1$ for $x \in R \setminus \{ -2 \}$,then $\int f(x) \,dx$ is equal to (where $C$ is a constant of integration)

If $f\left(\frac{t+1}{2 t+1}\right)=t+1$, then $\int f(x) d x=$

$\int {\{ 1 + 2\tan x(\tan x + \sec x)\} }^{1/2} dx = $