If $\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+c$,then find the value of $a$ (where $c$ is the constant of integration).

$\int \frac{2-\sin x}{2 \cos x+3} d x=$

The integral $\int \frac{dx}{(1 + \sqrt{x}) \cdot \sqrt{x} \sqrt{1 - x}}$ is equal to (where $c$ is a constant of integration)

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) = $

If $\int e^x \cos x \, dx = \frac{e^x}{2}(\cos x + \sin x)$ and $\int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx = \frac{f(x)}{24}[\cos (g(x)) + \sin (g(x))] + c$,then $g(1) =$

$\int \frac{3 \sin x+5 \cos x+4}{\sin x+\cos x+2} d x=$