$$\int_\limits1^3 x^n \sqrt{x^2-1} d x=6 \text {, then } n=$$

[ . ] represents greatest integer function, then $$\int_{-1}^1(x[1+\sin \pi x]+1) d x=$$

$$\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2 n+1}}+\frac{n}{(n+2) \sqrt{2(2 n+2)}}\right. \\ & \left.+\frac{n}{(n+3) \sqrt{3(2 n+3)}}+\ldots n \text { terms }\right]=\int_\limits0^1 f(x) d x \end{aligned}$$

then $$f(x)=$$

If $$I_n=\int_0^{\pi / 4} \tan ^n x d x$$, then $$\frac{1}{I_2+I_4}+\frac{1}{I_3+I_5}+\frac{1}{I_4+I_6}=$$