The value of $$\int\limits_0^1 {{{{x^4}{{\left( {1 - x} \right)}^4}} \over {1 + {x^2}}}dx} $$ is (are)

The value of $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int\limits_0^x {{{t\ln \left( {1 + t} \right)} \over {{t^4} + 4}}} dt$$ is

Let $$f$$ be a real-valued function defined on the interval $$(-1, 1)$$ such that
$${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt,} $$ for all $$x \in \left( { - 1,1} \right)$$,
and let $${f^{ - 1}}$$ be the inverse function of $$f$$. Then $$\left( {{f^{ - 1}}} \right)'\left( 2 \right)$$ is equal to

Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by
$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$