The value of the integral $$\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} $$ is

The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ is equal to :

Let $$f(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in R$$ be a function which

satisfies $$f(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy} $$. then $$(a+b)$$ is equal to

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