$\int \frac{\cos x-\sin x}{8-\sin 2x} dx = \frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$,then $p = \ldots$

If $f(x)+k$ is obtained by evaluating $\int \frac{x^3}{\left(1+x^2\right)^3} d x$ using the substitution $x=\tan \theta$, and $g(x)+c$ is obtained by evaluating $\int \frac{x^3}{\left(1+x^2\right)^3} d x$ using the substitution $x^2+1=z$, then $f(x)-g(x)+k-c=$

If $\int {\frac{{\log \left( {t + \sqrt {1 + {t^2}} } \right)}}{{\sqrt {1 + {t^2}} }}dt = \frac{1}{2}{{\left( {g\left( t \right)} \right)}^2} + C} $,where $C$ is a constant,then $g(2)$ is equal to

$\int \frac{3x^2}{x^6 + 1} dx = $

The integral $\int \frac{1}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}} dx$ is equal to : (where $C$ is a constant of integration)

If $\int_1^3 x^n \sqrt{x^2-1} dx = 6$,then $n = $