Let f(x) = $${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$$ x $$\, \in $$ R, where a, b and d are non-zero real constants. Then :

The integral  $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$  equals :

Let a function f : (0, $$\infty $$) $$ \to $$ (0, $$\infty $$) be defined by f(x) = $$\left| {1 - {1 \over x}} \right|$$. Then f is :

If   $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1} $$ + C, where C is a constant of integration, then f(x) is equal to :