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$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$$

$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 = - \frac{1}{{\sin t}} - \sin t+c$

$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}{{\left| {{\rm{co}}{{\rm{s}}^3}t} \right|}}$$\Rightarrow I = \int {\frac{{\frac{{dt}}{{{{\cos }^2}t}}}}{{\frac{{{{\sin }^2}t}}{{{{\cos }^2}t}}\frac{1}{{\left| {{\rm{co}}{{\rm{s}}^3}t} \right|}}}}} = \int {\frac{{\left| {{\rm{co}}{{\rm{s}}^3}t} \right|}}{{{{\sin }^2}t}}} dt = \int {\frac{{(1 - {{\sin }^2}t)\left| {{\rm{cos}}t} \right|}}{{{{\sin }^2}t}}} dt$\[ \Rightarrow I = \int {(\frac{1}{{{{\sin }^2}t}} - 1)} \left| {{\rm{cos}}t} \right|dt\]$*t \in ( - \frac{\pi }{2};0) \Rightarrow I = - \int {(\frac{1}{{{{\sin }^2}t}} - 1)} {\rm{cos}}tdt$$*t \in (0;\frac{\pi }{2}) \Rightarrow I = \int {(\frac{1}{{{{\sin }^2}t}} - 1)} {\rm{cos}}tdt$Với mỗi trường hợp, đặt tiếp $u = \sin t$ là xong.