The parametric form of a curve is $$x=\frac{t^3}{t^2-1} y=\frac{t}{t^2-1}$$, then $$\int \frac{d x}{x-3 y}=$$

Given, $$\frac{3 x-2}{(x+1)^2(x+3)}=\frac{A}{x+1} +\frac{B}{(x+1)^2}+\frac{C}{x+3}$$, then $$4 A+2 B+4 C$$ is equal to

$$\int \frac{\sin \alpha}{\sqrt{1+\cos \alpha}} d \alpha$$ is equal to

If $$\int \frac{\cos 4 x+1}{\cot x-\tan x}=k \cos 4 x+C$$, then $$k$$ is equal to