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

If $n(2n+1) \int_{0}^{1}(1-x^n)^{2n} dx = 1177 \int_{0}^{1}(1-x^n)^{2n+1} dx$,then $n \in N$ is equal to $\dots\dots$

Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

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:

Let $f: R \rightarrow R$ be a function defined as $f(x) = a \sin \left(\frac{\pi[x]}{2}\right) + [2-x]$,$a \in R$,where $[t]$ is the greatest integer less than or equal to $t$. If $\lim_{x \rightarrow -1} f(x)$ exists,then the value of $\int_{0}^{4} f(x) dx$ is equal to.

Let $f$ be a real-valued continuous function on $[0, 1]$ and $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$. Then which of the following points $(x, y)$ lies on the curve $y = f(x)$?