If $I_n = \int \sin^n x \, dx$, then $n I_n - (n-1) I_{n-2}$ equals

For any integer $n \geq 2$, let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ 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 $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} =$

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx = $

If $\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx = g(x) + c$ and $g(1) = 0$,then $g\left(\frac{1}{2}\right)$ is equal to

If $\int(1+x) \log \left(1+x^2\right) d x=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \log \left(1+x^2\right)+g(x)+C$,then $g(x)=$