$\int \sqrt{1 + x^2} \, dx = $

If $\int {\frac{{\csc^2 x}}{{{{\left( {\csc x + \cot x} \right)}^{\frac{9}{2}}}}}\,dx} = {\left( {\csc x - \cot x} \right)^{\frac{7}{2}}}\left( {\frac{1}{\alpha } + \frac{{{{\left( {\csc x - \cot x} \right)}^2}}}{{11}}} \right) + C$ (where $C$ is the constant of integration and $\alpha \in N$),then $\alpha$ is:

$\int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx = A \{23x + 2\ln |4\sin x + 5\cos x|\} + c$,then $A$ and $f(x)$ are

If $\int \frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2} d x=f(x)+c$, then $f(x)$ is equal to

$\int \frac{dx}{7+5 \cos x}$ is equal to

$\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x=$