If $I_n = \int_{0}^{1} \frac{dx}{(1 + x^2)^n}$; $n \in N$,then which of the following statements hold good?

Let $I_1 = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx$,$I_2 = \int_0^{2\pi} \cos^6 x dx$,$I_3 = \int_{-\pi/2}^{\pi/2} \sin^3 x dx$,and $I_4 = \int_0^1 \ln \left( \frac{1}{x} - 1 \right) dx$. Then:

An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

If $I_1 = \int_0^{\pi / 2} \frac{x}{\sin x} dx$ and $I_2 = \int_0^1 \frac{\tan^{-1} x}{x} dx$,then $I_1 : I_2$ is

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.

The number of solutions of the equation $6 \int_{0}^{|x|} ((t^2-1) \ln t) dt = 5|x|$ for $x \in R \setminus \{0\}$ is