Let $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$ and $v = \int_{0}^{\infty} \frac{x^2 dx}{x^4 + 7x^2 + 1}$. Then:

Let $\{x\}$ and $[x]$ denote the fractional part of $x$ and the greatest integer $\leq x$ respectively of a real number $x$. If $\int_{0}^{n}\{x\} dx$,$\int_{0}^{n}[x] dx$,and $10(n^{2}-n)$ $(n \in N, n > 1)$ are three consecutive terms of a $G.P.$,then $n$ is equal to

Let the function $f: R \rightarrow R$ be defined by
$f(t)=\begin{cases} (-1)^{n+1} 2, & \text{if } t=2n-1, n \in N \\ \frac{(2n+1-t)}{2} f(2n-1) + \frac{(t-(2n-1))}{2} f(2n+1), & \text{if } 2n-1 < t < 2n+1, n \in N \end{cases}$
Define $g(x) = \int_1^x f(t) dt, x \in (1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x) = 0$ in the interval $(1, 8]$ and $\beta = \lim_{x \rightarrow 1^+} \frac{g(x)}{x-1}$. Then the value of $\alpha + \beta$ is equal to.

Let $I_{n} = \int_{0}^{1} x^{n} \tan^{-1} x \, dx$. If $a_{n} I_{n+2} + b_{n} I_{n} = c_{n}$ for all $n \geq 1$, then

If the abscissa of the vertex of the parabola $y = ax^2 + bx + c$ is $1$ $(a, b, c > 0)$ and $f(x) = \int_0^x (3at^2 + bt + c) dt$ is a strictly increasing function $\forall x \in R$,then the maximum possible value of $[\frac{a}{c}]$ is (where $[.]$ denotes the greatest integer function).

If $\alpha = \int_{0}^{2\sqrt{3}} \log_2(x^2+4) dx + \int_{2}^{4} \sqrt{2^x-4} dx$, then $\alpha^2$ is equal to . . . . . . .