The value of $\int {\frac{{2dx}}{{\sqrt {1 - 4{x^2}} }}} $ is

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

$\int \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx = $

$\int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \,d x=$

$\int \frac{dx}{5 + 4\cos x} = $

If $\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$,then $\frac{P}{Q} =$