K
kachia_17
[tex]\huge \blue I_8=\int\frac{dx}{1+x^8} \\ \ \\ \Leftrightarrow I=\int\frac{1+\sqrt2.x^2+x^4-(1+\sqrt2.x^2+x^4)}{2\sqrt2 x^2(1+\sqrt2.x^2+x^4)(1-\sqrt2.x^2+x^4)} \\ \ \\ \Leftrightarrow I=\int\frac{1+\sqrt2.x^2+x^4}{2\sqrt2 x^2(1+\sqrt2.x^2+x^4)(1-\sqrt2.x^2+x^4)}dx-\int\frac{1-\sqrt2.x^2+x^4}{2\sqrt2 x^2(1+\sqrt2.x^2+x^4)(1-\sqrt2.x^2+x^4)}dx \\ \ \\ \Leftrightarrow I=\int\frac{dx}{2\sqrt2 x^2(1-\sqrt2.x^2+x^4)}-\int\frac{dx}{2\sqrt2 x^2(1+\sqrt2.x^2+x^4)} \\ \ \\ \Leftrightarrow I=J_a - J_b \\ J_a=\int\frac{dx}{2\sqrt2 x^2(1-\sqrt2.x^2+x^4)} \\ \ \\ \Leftrightarrow J_a=\int\frac{(1-\sqrt2.x^2+x^4)+\sqrt2.x^2-x^4}{2\sqrt2.x^2(1-\sqrt2.x^2+x^4)} \\ \ \\ \Leftrightarrow J_a=\int\frac{dx}{2\sqrt2.x^2}+\int\frac{dx}{2(1-\sqrt2.x^2+x^4)}-\int\frac{x^2}{2\sqrt2(1-\sqrt2.x^2+x^4}dx \\ \ \\ \Leftrightarrow J_a=-\frac{1}{\sqrt2.x^3} -\frac{1}{2\sqrt2} .\int\frac{x^2-\sqrt2}{1-\sqrt2.x^2+x^4}dx \\ \ \\ \Leftrightarrow J_a=-\frac{1}{\sqrt2.x^3}-\frac{1}{2\sqrt2}.K_a \\ \huge K_a = \int {\frac{{x^2 - \sqrt 2 }}{{1 - \sqrt 2.x^2+x^4}}dx}=\int{\frac{{x^2 - 1 + 1 - \sqrt 2 }}{{1 - \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1 - \left( {x^2 - 1} \right)} \right)}}{{1 - \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \int {\frac{{\left( {x^2 - 1} \right) - \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1} \right)}}{{1 - \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{\left( {\frac{{1 + \sqrt 2 }}{2}} \right)\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1} \right)}}{{1 - \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{x^2 - 1}}{{1 - \sqrt 2 .x^2 + x^4 }}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{x^2 + 1}}{{1 - \sqrt 2 .x^2 + x^4 }}dx} =\\ {\rm{ }} = \left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{1 - \frac{1}{{x^2 }}}}{{\frac{1}{{x^2 }} + x^2 - \sqrt 2 }}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{1 + \frac{1}{{x^2 }}}}{{\frac{1}{{x^2 }} + x^2 - \sqrt 2 }}dx} =\\{\rm{ = }}\left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{d\left( {x + \frac{1}{x}} \right)}}{{\left( {\frac{1}{x} + x} \right)^2 - \left( {2 + \sqrt 2 } \right)}}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{d\left( {x - \frac{1}{x}} \right)}}{{\left( {x - \frac{1}{x}} \right)^2 + \left( {2 - \sqrt 2 } \right)}}dx}[/tex]
[tex]\huge \blue J_b = \int {\frac{{dx}}{{2\sqrt 2 .x^2 .\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}}} = \int {\frac{{\left[ {1 + \sqrt 2 .x^2 + x^4 } \right] - \sqrt 2 .x^2 - x^4 }}{{2\sqrt 2 .x^2 .\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}}dx} = \\ {\rm{ }} = \int {\frac{{dx}}{{2\sqrt 2 .x^2 }} + \int {\frac{{dx}}{{2.\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}} - \int {\frac{{x^2 }}{{2\sqrt 2 .\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}}dx} } } = \\ {\rm{ }} = - \frac{1}{{\sqrt 2 .x^3 }} - \frac{1}{{2\sqrt 2 }}.\int {\frac{{x^2 - \sqrt 2 }}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = - \frac{1}{{\sqrt 2 .x^3 }} - \frac{1}{{2\sqrt 2 }}.K_b \\ \ \\ \ \\ \ K_b = \int {\frac{{x^2 - \sqrt 2 }}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{x^2 - 1 + 1 - \sqrt 2 }}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1 - \left( {x^2 - 1} \right)} \right)}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \int {\frac{{\left( {x^2 - 1} \right) - \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1} \right)}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{\left( {\frac{{1 + \sqrt 2 }}{2}} \right)\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1} \right)}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{x^2 - 1}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{x^2 + 1}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{1 - \frac{1}{{x^2 }}}}{{\frac{1}{{x^2 }} + x^2 + \sqrt 2 }}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{1 + \frac{1}{{x^2 }}}}{{\frac{1}{{x^2 }} + x^2 + \sqrt 2 }}dx} = \\ {\rm{ = }}\left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{d\left( {x + \frac{1}{x}} \right)}}{{\left( {\frac{1}{x} + x} \right)^2 - \left( {2 - \sqrt 2 } \right)}}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{d\left( {x - \frac{1}{x}} \right)}}{{\left( {x - \frac{1}{x}} \right)^2 + \left( {2 + \sqrt 2 } \right)}}dx} [/tex]
Đây là 1 bài khó ! Nếu chỉ phục vụ thi đại học thì không nên đâm đầu vào mấy dạng này làm gì !Chủ đề này sẽ được khóa lại, ai có nhu cầu thắc mắc gì thì liên hệ kachia !
[tex]\huge \blue J_b = \int {\frac{{dx}}{{2\sqrt 2 .x^2 .\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}}} = \int {\frac{{\left[ {1 + \sqrt 2 .x^2 + x^4 } \right] - \sqrt 2 .x^2 - x^4 }}{{2\sqrt 2 .x^2 .\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}}dx} = \\ {\rm{ }} = \int {\frac{{dx}}{{2\sqrt 2 .x^2 }} + \int {\frac{{dx}}{{2.\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}} - \int {\frac{{x^2 }}{{2\sqrt 2 .\left[ {1 + \sqrt 2 .x^2 + x^4 } \right]}}dx} } } = \\ {\rm{ }} = - \frac{1}{{\sqrt 2 .x^3 }} - \frac{1}{{2\sqrt 2 }}.\int {\frac{{x^2 - \sqrt 2 }}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = - \frac{1}{{\sqrt 2 .x^3 }} - \frac{1}{{2\sqrt 2 }}.K_b \\ \ \\ \ \\ \ K_b = \int {\frac{{x^2 - \sqrt 2 }}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{x^2 - 1 + 1 - \sqrt 2 }}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1 - \left( {x^2 - 1} \right)} \right)}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \int {\frac{{\left( {x^2 - 1} \right) - \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1} \right)}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \int {\frac{{\left( {\frac{{1 + \sqrt 2 }}{2}} \right)\left( {x^2 - 1} \right) + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\left( {x^2 + 1} \right)}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{x^2 - 1}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{x^2 + 1}}{{1 + \sqrt 2 .x^2 + x^4 }}dx} = \\ {\rm{ }} = \left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{1 - \frac{1}{{x^2 }}}}{{\frac{1}{{x^2 }} + x^2 + \sqrt 2 }}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{1 + \frac{1}{{x^2 }}}}{{\frac{1}{{x^2 }} + x^2 + \sqrt 2 }}dx} = \\ {\rm{ = }}\left( {\frac{{1 + \sqrt 2 }}{2}} \right).\int {\frac{{d\left( {x + \frac{1}{x}} \right)}}{{\left( {\frac{1}{x} + x} \right)^2 - \left( {2 - \sqrt 2 } \right)}}dx} + \left( {\frac{{1 - \sqrt 2 }}{2}} \right).\int {\frac{{d\left( {x - \frac{1}{x}} \right)}}{{\left( {x - \frac{1}{x}} \right)^2 + \left( {2 + \sqrt 2 } \right)}}dx} [/tex]
Đây là 1 bài khó ! Nếu chỉ phục vụ thi đại học thì không nên đâm đầu vào mấy dạng này làm gì !Chủ đề này sẽ được khóa lại, ai có nhu cầu thắc mắc gì thì liên hệ kachia !
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