Tích phân được viết thành:
\[I = \int\limits_{\frac{1}{{\sqrt 3 }}}^1 {\frac{{dt}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}}} + \int\limits_1^{\sqrt 3 } {\frac{{dt}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}} = {I_1} + {I_2}} \]
Tính tích phân: ${I_1} = \int\limits_{\frac{1}{{\sqrt 3 }}}^1 {\frac{{dx}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}}} $
Đặt $u = \frac{1}{t} \Rightarrow dt = - \frac{{du}}{{{u^2}}};\,\,t:\frac{1}{{\sqrt 3 }} \to 1 \Rightarrow u:\sqrt 3 \to 1$
Khi đó: \[{I_1} = \int\limits_{\sqrt 3 }^1 {\frac{1}{{\left( {1 + \frac{1}{{{u^2}}}} \right)\left( {1 + \frac{1}{{{u^{\sqrt 2 }}}}} \right)}}} \left( { - \frac{1}{{{u^2}}}} \right)du = \int\limits_1^{\sqrt 3 } {\frac{{{u^{\sqrt 2 }}}}{{\left( {1 + {u^2}} \right)\left( {1 + {u^{\sqrt 2 }}} \right)}}du = } \int\limits_1^{\sqrt 3 } {\frac{{{t^{\sqrt 2 }}}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}}dt} \]
Do đó: \[I = \int\limits_1^{\sqrt 3 } {\frac{{{t^{\sqrt 2 }}}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}}dt} + \int\limits_1^{\sqrt 3 } {\frac{{dt}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}}} = \int\limits_1^{\sqrt 3 } {\frac{{\left( {1 + {t^{\sqrt 2 }}} \right)dt}}{{\left( {1 + {t^2}} \right)\left( {1 + {t^{\sqrt 2 }}} \right)}}} \]
\[ = \int\limits_1^{\sqrt 3 } {\frac{{dt}}{{1 + {t^2}}}} = \left. {arctg\left( t \right)} \right|_1^{\sqrt 3 } = arctg\sqrt 3 - arctg1 = \boxed{\dfrac{\pi }{{12}}}\]
