Answer
see below answer
Work Step by Step
a)
let
$x$ = ${\sqrt 2}secθ$
$dx$ = ${\sqrt 2}secθtanθdθ$
and
$\sqrt {x^{2}-2}$ = $\sqrt {2sec^{2}θ-2}$ = $\sqrt {2(sec^{2}θ-1)}$ = $\sqrt {2tan^{2}θ}$ = $\sqrt 2{tanθ}$
so
$I$ = $\frac{dx}{x^{2}\sqrt {x^{2}-2}}$ = $\int{\frac{\sqrt 2{}secθtanθdθ}{(2sec^{2}θ)(\sqrt {2}tanθ)}}$ = $\frac{1}{2}\int{\frac{dθ}{secθ}}$ = $\frac{1}{2}\int{cosθ}dθ$ = $\frac{1}{2}sinθ+C$
b)
$x$ = ${\sqrt 2}secθ$
$cosθ$ = $\frac{\sqrt 2}{x}$
as file attached
$sinθ$ = $\frac{\sqrt {x^{2}-2}}{x}$
c)
$I$ = $\frac{1}{2}sinθ+C$ = $\frac{\sqrt {x^{2}-2}}{2x}+C$
