$I = \int {\frac{{dx}}{{{x^2}\sqrt {{{(1 + {x^2})}^3}} }}} $Đặt $x = \tan t,t \in ( - \frac{\pi }{2};0) \cup (0;\frac{\pi }{2})$
$\Rightarrow dx = \frac{{dt}}{{{{\cos }^2}t}}$
$\sqrt {{{(1 + {x^2})}^3}} = \sqrt {{{(1 + {{\tan }^2}t)}^3}} = {\left( {\sqrt {\frac{1}{{{{\cos }^2}t}}} } \right)^3} = \frac{1}{{{\rm{co}}{{\rm{s}}^3}t}}$ (do $t \in ( - \frac{\pi }{2};0) \cup (0;\frac{\pi }{2})$)
$\Rightarrow I = \int {\frac{{\frac{{dt}}{{{{\cos }^2}t}}}}{{\frac{{{{\sin }^2}t}}{{{{\cos }^2}t}}\frac{1}{{{\rm{co}}{{\rm{s}}^3}t}}}}} = \int {\frac{{{\rm{co}}{{\rm{s}}^3}t}}{{{{\sin }^2}t}}} dt = \int {\frac{{(1 - {{\sin }^2}t){\rm{cos}}t}}{{{{\sin }^2}t}}} dt$
\[ \Rightarrow I = \int {(\frac{1}{{{{\sin }^2}t}} - 1)} {\rm{cos}}tdt\]
Đặt $u = \sin t \Rightarrow du = \cos tdt$
$ \Rightarrow I = \int {(\frac{1}{{{u^2}}} - 1)} du = - \frac{1}{u} - u + c$
Ta có:
$x = \tan t = \frac{{\sin t}}{{\cos t}} = \frac{u}{{\sqrt {1 - {u^2}} }}$
$ \Rightarrow {x^2}(1 - {u^2}) = {u^2}$
$\Rightarrow u = \frac{x}{{\sqrt {1 + {x^2}} }}$
Vậy: $$I = - \frac{{\sqrt {1 + {x^2}} }}{x} - \frac{x}{{\sqrt {1 + {x^2}} }} + c$$