The value of $\int\limits_0^2 {\frac{{dx}}{{{{(1 - x)}^2}}}} $ is

If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

$A$ continuous and differentiable function $f$ satisfies the condition $\int_{0}^{x} f(t) dt = f^2(x) - 1$ for all real $x$. Then:

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 $I_n = \int_0^1 (\log x)^n dx$,where $n$ is a non-negative integer. Then,$I_{2011} + 2011 I_{2010}$ is equal to

The value of $\int\limits_0^{\frac{\pi }{2}} {\sin 8x \cot x \, dx} + \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\ln \left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)dx}$ is equal to