$\int\limits_1^e {\left( {\frac{{{{\tan }^{ - 1}}x}}{x} + \frac{{\ln x}}{{1 + {x^2}}}} \right)} \,dx$ is equal to

$\int_0^1 {{\cos }^{ - 1}}x\,dx = $

If $f : R \to R$ is a continuous function such that $f(x) = \int\limits_1^x {tf(t)dt}$,then the correct statement is -

The set of values of $a$ which satisfy the equation $\int_{0}^{2} (t - \log_{2} a) \, dt = \log_{2} \left( \frac{4}{a^{2}} \right)$ is

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

If $\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}$,where $a, b, c$ are rational numbers,then $2 a+3 b-4 c$ is equal to :