The integral $$16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}} $$ is equal to

The minimum value of the function $$f(x) = \int\limits_0^2 {{e^{|x - t|}}dt} $$ is :

$$\int_{{{3\sqrt 2 } \over 4}}^{{{3\sqrt 3 } \over 4}} {{{48} \over {\sqrt {9 - 4{x^2}} }}dx} $$ is equal to

The area enclosed by the curves $${y^2} + 4x = 4$$ and $$y - 2x = 2$$ is :