$\int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx =$

If $\int \left( \frac{4 e^x + 6 e^{-x}}{9 e^x - 4 e^{-x}} \right) d x = A x + B \log |9 e^{2 x} - 4| + C$,then $(A, B) = $

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).

$I_n = \int \frac{t^n}{1+t^2} dt, (n = 1, 2, 3, \ldots) \Rightarrow I_6 + I_4 =$

If $\int \frac{2 x^2+5 x+9}{\sqrt{x^2+x+1}} d x=x \sqrt{x^2+x+1}+\alpha \sqrt{x^2+x+1}+\beta \log _{ e }\left| x +\frac{1}{2}+\sqrt{ x ^2+ x +1}\right|+ C$,where $C$ is the constant of integration,then $\alpha+2 \beta$ is equal to . . . . . .

If $\int \frac{1}{x^4+8 x^2+9} d x = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right] + c$, then $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} =$