Sửa lại bài này
Cách 1( Dùng cho Phổ thông):
[TEX]I = \int {{{dx} \over {x^8 + 1}}}[/TEX]
Ta có: [TEX]{1 \over {x^8 + 1}} = {1 \over {(x^4 - x^2 \sqrt 2 + 1)(x^4 + x^2 \sqrt 2 + 1)}} = {{A_1 x^3 + B_1 x^2 + C_1 x + D_1 } \over {(x^4 - x^2 \sqrt 2 + 1)}} + {{A_2 x^3 + B_2 x^2 + C_2 x + D_2 } \over {(x^4 + x^2 \sqrt 2 + 1)}}[/TEX]
Dùng phương pháp hệ số bất định ta có:
[TEX]\int {{{dx} \over {x^8 + 1}}} = \int {{{ - {1 \over {2\sqrt 2 }}x^2 + {1 \over 2}} \over {(x^4 - x^2 \sqrt 2 + 1)}}} dx + \int {{{{1 \over {2\sqrt 2 }}x^2 + {1 \over 2}} \over {(x^4 + x^2 \sqrt 2 + 1)}}} dx = I_1 + I_2 [/TEX].
* Tính I_1: [TEX]I_1 = \int {{{ - {1 \over {2\sqrt 2 }}x^2 + {1 \over 2}} \over {x^4 - x^2 \sqrt 2 + 1}}} dx = - {1 \over {2\sqrt 2 }}\int {{{x^2 + 1} \over {x^4 - x^2 \sqrt 2 + 1}}} dx + ({1 \over 2} + {1 \over {2\sqrt 2 }})\int {{{dx} \over {x^4 - x^2 \sqrt 2 + 1}}} = J_1 + J_2[/TEX].
[TEX] J_1 = - {1 \over {2\sqrt 2 }}\int {{{x^2 + 1} \over {x^4 - x^2 \sqrt 2 + 1}}} dx = - {1 \over {2\sqrt 2 }}\int {{{1 + {1 \over {x^2 }}} \over {x^2 - \sqrt 2 + {1 \over {x^2 }dx = - {1 \over {2\sqrt 2 }}\int {{{d(x - {1 \over x})} \over {(x - {1 \over x})^2 + (2 - \sqrt 2 )}}} }= - {1 \over {2\sqrt {4 - 2\sqrt 2 } }}{\rm{ar}}ctg{{(x - {1 \over x})} \over {\sqrt {2 - \sqrt 2 } }} + C_1 \cr}[/TEX]
[TEX]J_2 = ({1 \over 2} + {1 \over {2\sqrt 2 }})\int {{{dx} \over {x^4 - x^2 \sqrt 2 + 1}}} = ({1 \over 2} + {1 \over {2\sqrt 2 }})\int {{{dx} \over {(x^2 - {1 \over {\sqrt 2 }})^2 + {1 \over 2}}}} = {1 \over {\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }})\int {{{dt} \over {\sqrt {tg\,t + 1} }}} \,\,\,\,\,\,\,\,\,\left( {{1 \over {\sqrt 2 }}tg\,t = x^2 - {1 \over {\sqrt 2 }} \Rightarrow dx = {{dt} \over {2\root 4 \of 2 c{\rm{os}}^2 t\sqrt {tg\,t + 1} }}} \right) = {1 \over {\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }})\int {\left( {{1 \over {2u}} - {1 \over 4}{{2(u - 1)} \over {u^2 - 2u + 2}} + {1 \over 2}{1 \over {(u - 1)^2 + 1}}} \right)} du\,\,\,\,\,\,\,\,\left( {u = tg\,t + 1 \Rightarrow dt = {{du} \over {u^2 - 2u + 2}}} \right) = {1 \over {4\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }})\ln {{u^2 } \over {u^2 - 2u + 2}} + ({1 \over 2} + {1 \over {2\sqrt 2 }}){1 \over {2\root 4 \of 2 }}{\rm{ar}}ctg(u - 1) = {1 \over {4\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }})\ln {{2x^4 } \over {2x^4 - 2\sqrt 2 x^2 + 2}} + {1 \over {2\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }}){\rm{ar}}ctg(\sqrt 2 x^2 - 1) \cr}[/TEX]
[TEX]I_1 = - {1 \over {4\sqrt {4 - 2\sqrt 2 } }}{\rm{ar}}ctg\left( {{{x - {1 \over x}} \over {\sqrt {2 - \sqrt 2 } }}} \right) + {1 \over {4\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }})\ln {{2x^4 } \over {2x^4 - 2\sqrt 2 x^2 + 2}} + {1 \over {2\root 4 \of 2 }}({1 \over 2} + {1 \over {2\sqrt 2 }}){\rm{ar}}ctg(\sqrt 2 x^2 - 1) + C \cr}[/TEX]
* Tính I_2:
[TEX]I_2 = \int {{{{1 \over {2\sqrt 2 }}x^2 + {1 \over 2}} \over {x^4 + x^2 \sqrt 2 + 1}}} dx = {1 \over {2\sqrt 2 }}\int {{{x^2 + 1} \over {x^4 + x^2 \sqrt 2 + 1}}} dx + ({1 \over 2} - {1 \over {2\sqrt 2 }})\int {{{dx} \over {x^4 + x^2 \sqrt 2 + 1}}} = K_1 + K_2[/TEX]
[TEX] K_1 = {1 \over {2\sqrt 2 }}\int {{{x^2 + 1} \over {x^4 + x^2 \sqrt 2 + 1}}} dx = {1 \over {2\sqrt 2 }}\int {{{1 + {1 \over {x^2 }}} \over {x^2 + \sqrt 2 + {1 \over {x^2 }}}}dx = {1 \over {2\sqrt 2 }}\int {{{d(x - {1 \over x})} \over {(x - {1 \over x})^2 + (2 + \sqrt 2 )}}}= {1 \over {2\sqrt {4 + 2\sqrt 2 } }}{\rm{ar}}ctg{{(x - {1 \over x})} \over {\sqrt {2 + \sqrt 2 } }} + C_2[/TEX]
[TEX]K_2 = ({1 \over 2} - {1 \over {2\sqrt 2 }})\int {{{dx} \over {x^4 + x^2 \sqrt 2 + 1}}} = ({1 \over 2} - {1 \over {2\sqrt 2 }})\int {{{dx} \over {(x^2 + {1 \over {\sqrt 2 }})^2 + {1 \over 2}}}} = {1 \over {\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }})\int {{{dt} \over {\sqrt {tg\,t + 1} }}} \,\,\,\,\,\,\,\,\,\left( {{1 \over {\sqrt 2 }}tg\,t = x^2 + {1 \over {\sqrt 2 }} \Rightarrow dx = {{dt} \over {2\root 4 \of 2 c{\rm{os}}^2 t\sqrt {tg\,t + 1} }}} \right) = {1 \over {\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }})\int {\left( {{1 \over {2u}} - {1 \over 4}{{2(u - 1)} \over {u^2 - 2u + 2}} + {1 \over 2}{1 \over {(u - 1)^2 + 1}}} \right)} du,\left( {u = tg\,t + 1 \Rightarrow dt = {{du} \over {u^2 - 2u + 2}}} \right) = {1 \over {4\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }})\ln {{u^2 } \over {u^2 - 2u + 2}} + {1 \over {2\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }}){\rm{ar}}ctg(u - 1) = {1 \over {4\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }})\ln {{(x^2 \sqrt 2 + 2)^2 } \over {2x^4 + 2\sqrt 2 x^2 + 2}} + {1 \over {2\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }}){\rm{ar}}ctg(\sqrt 2 x^2 + 1)[/TEX]
[TEX] I_2 = {1 \over {2\sqrt {4 + 2\sqrt 2 } }}{\rm{ar}}ctg{{(x - {1 \over x})} \over {\sqrt {2 + \sqrt 2 } }} + {1 \over {4\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }})\ln {{(x^2 \sqrt 2 + 2)^2 } \over {2x^4 + 2\sqrt 2 x^2 + 2}} + {1 \over {2\root 4 \of 2 }}({1 \over 2} - {1 \over {2\sqrt 2 }}){\rm{ar}}ctg(\sqrt 2 x^2 + 1)[/TEX]
Kết luận: I=I1+I2
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
