If $\int e^{3x} \cos 4x \,dx = e^{3x} (A \sin 4x + B \cos 4x) + c$,then:

Given that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. If $\int \frac{1}{x^4+3x^2+1} dx = a \cdot \tan^{-1}\left(\frac{b(x^2-1)}{x}\right) + c \cdot \tan^{-1}\left(\frac{d(x^2+1)}{x}\right) + k$, where $k$ is a constant of integration, then $5(c+d+ab) = $

$\int \frac{3x+2}{4x^2+4x+5} dx = A \log(4x^2+4x+5) + B \tan^{-1}\left(\frac{2x+1}{2}\right) + c$, then $A+B=$

$\int \sqrt{\frac{a+x}{a-x}} \, dx = $

Let $\int {{\sec }^{ - 1}}\left[ { - {\sin }^2x} \right]dx = f(x) + C$,(valid for $x \neq 0$) where $[k]$ denotes the greatest integer less than or equal to $k$ and $f(0) = 0$. Then the value of ${\left( {f\left( {\frac{8}{{\pi x}}} \right)} \right)''}$ at $x = 2$ is (where $'$ denotes the derivative).

If $\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\cot x \sin \theta}+C,$ where $C$ is the integration constant,then $AB$ is equal to