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Question-137037

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Question Number 137037 by JulioCesar last updated on 29/Mar/21
Answered by mathmax by abdo last updated on 29/Mar/21
I=∫ (dx/((x^3 −8)^2 )) (compolex method) ⇒I=∫ (dx/((x−2)^2 (x^2 +2x+4)^2 )) x^2 +2x+4=0→Δ^′ =−3 ⇒x_1 =−1+i(√3)=2(−(1/2)+((i(√3))/2))=2e^((2iπ)/3) x_2 =−1−i(√3)=2e^(−((2iπ)/3)) ⇒I=∫ (dx/((x−2)^2 (x−2e^((2iπ)/3) )^2 (x−e^(−((2iπ)/3)) ))) =∫ (dx/((x−2)^2 (((x−e^((2iπ)/3) )/(x−e^(−((2iπ)/3)) )))^2 (x−e^(−((2iπ)/3)) )^4 )) changement ((x−e^((2iπ)/3) )/(x−e^(−((2iπ)/3)) ))=z ⇒x−e^((2iπ)/3) =zx−e^(−((2iπ)/3)) z ⇒ (1−z)x=e^((2iπ)/3) −e^(−((2iπ)/3)) z ⇒x =((e^(−((2iπ)/3)) z−e^((2iπ)/3) )/(z−1)) ⇒ (dx/dz)=((e^(−((2iπ)/3)) (z−1)−e^(−((2iπ)/3)) z+e^((2iπ)/3) )/((z−1)^2 ))=((2isin((π/3)))/((z−1)^2 ))=((2i((√3)/2))/((z−1)^2 ))=((i(√3))/((z−1)^2 )) x−2=((e^(−((2iπ)/3)) z−e^((2iπ)/3) )/(z−1))−2 =((e^(−((2iπ)/3)) z−e^((2iπ)/3) −2z+2)/(z−1)) x−e^(−((2iπ)/3)) =((e^(−((2iπ)/3)) z−e^((2iπ)/3) )/(z−1))−e^((−2iπ)/3) =((−e^((2iπ)/3) +e^(−((2iπ)/3)) )/(z−1)) =((−i(√3))/((z−1))) ⇒ I = ∫ (1/(((((e^(−((2iπ)/3)) −2)z+2−e^((2iπ)/3) )/(z−1)))^2 .z^2 (((−i(√3))/(z−1)))^4 ))((i(√3))/((z−1)^2 ))dz =(1/((i(√3))^3 ))∫ (((z−1)^4 )/(z^2 (αz +β)^2 ))dz (α=(e^(−((2iπ)/3)) −2) and β=2−e^((2iπ)/3) ) but ∫ (((z−1)^4 )/(z^2 (αz+β)^2 ))dz =∫ (1/(z^2 (αz+β)^2 ))Σ_(k=0) ^4 C_4 ^k z^k (−1)^(4−k) dz =Σ_(k0) ^4 C_4 ^k (−1)^k ∫ (z^(k−2) /((αz+β)^2 ))dz =_(αz+β=t) Σ_(k=0) ^4 C_4 ^k (−1)^k ∫ (((((t−β)/α))^(k−2) )/t^2 )(dt/α) =Σ_(k=0) ^4 (−1)^k C_4 ^k (1/α^(k−1) ) ∫ (((t−β)^(k−2) )/t^2 )dt and I_k =∫ (((t−β)^(k−2) )/t^2 )dt are eazy to find....
$$\mathrm{I}=\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{8}\right)^{\mathrm{2}} }\:\left(\mathrm{compolex}\:\mathrm{method}\right)\:\Rightarrow\mathrm{I}=\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{4}=\mathrm{0}\rightarrow\Delta^{'} =−\mathrm{3}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{x}_{\mathrm{2}} =−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\Rightarrow\mathrm{I}=\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{2e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)} \\ $$$$=\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} \left(\frac{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} }\right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)^{\mathrm{4}} } \\ $$$$\mathrm{changement}\:\frac{\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} }=\mathrm{z}\:\Rightarrow\mathrm{x}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} =\mathrm{zx}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\mathrm{z}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{z}\right)\mathrm{x}=\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} −\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \mathrm{z}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{z}−\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dz}}=\frac{\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \left(\mathrm{z}−\mathrm{1}\right)−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \mathrm{z}+\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2isin}\left(\frac{\pi}{\mathrm{3}}\right)}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{x}−\mathrm{2}=\frac{\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{z}−\mathrm{1}}−\mathrm{2}\:=\frac{\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} −\mathrm{2z}+\mathrm{2}}{\mathrm{z}−\mathrm{1}} \\ $$$$\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:=\frac{\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \mathrm{z}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{z}−\mathrm{1}}−\mathrm{e}^{\frac{−\mathrm{2i}\pi}{\mathrm{3}}} \:=\frac{−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} +\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{z}−\mathrm{1}}\:=\frac{−\mathrm{i}\sqrt{\mathrm{3}}}{\left(\mathrm{z}−\mathrm{1}\right)}\:\Rightarrow \\ $$$$\mathrm{I}\:=\:\int\:\:\:\:\frac{\mathrm{1}}{\left(\frac{\left(\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} −\mathrm{2}\right)\mathrm{z}+\mathrm{2}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} }{\mathrm{z}−\mathrm{1}}\right)^{\mathrm{2}} .\mathrm{z}^{\mathrm{2}} \left(\frac{−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{z}−\mathrm{1}}\right)^{\mathrm{4}} }\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\int\:\:\:\:\frac{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{z}^{\mathrm{2}} \left(\alpha\mathrm{z}\:+\beta\right)^{\mathrm{2}} }\mathrm{dz}\:\:\:\:\:\:\:\left(\alpha=\left(\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} −\mathrm{2}\right)\:\mathrm{and}\:\beta=\mathrm{2}−\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)\:\mathrm{but} \\ $$$$\int\:\:\frac{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{z}^{\mathrm{2}} \left(\alpha\mathrm{z}+\beta\right)^{\mathrm{2}} }\mathrm{dz}\:=\int\:\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} \left(\alpha\mathrm{z}+\beta\right)^{\mathrm{2}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \:\mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \:\mathrm{z}^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{4}−\mathrm{k}} \:\mathrm{dz} \\ $$$$=\sum_{\mathrm{k0}} ^{\mathrm{4}} \:\mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \int\:\frac{\mathrm{z}^{\mathrm{k}−\mathrm{2}} }{\left(\alpha\mathrm{z}+\beta\right)^{\mathrm{2}} }\mathrm{dz} \\ $$$$=_{\alpha\mathrm{z}+\beta=\mathrm{t}} \:\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \:\mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \:\int\:\:\frac{\left(\frac{\mathrm{t}−\beta}{\alpha}\right)^{\mathrm{k}−\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }\frac{\mathrm{dt}}{\alpha} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \frac{\mathrm{1}}{\alpha^{\mathrm{k}−\mathrm{1}} }\:\int\:\:\:\frac{\left(\mathrm{t}−\beta\right)^{\mathrm{k}−\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:\mathrm{and} \\ $$$$\mathrm{I}_{\mathrm{k}} =\int\:\:\frac{\left(\mathrm{t}−\beta\right)^{\mathrm{k}−\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\mathrm{are}\:\mathrm{eazy}\:\mathrm{to}\:\mathrm{find}…. \\ $$
Answered by MJS_new last updated on 30/Mar/21
∫(dx/((x^3 −8)^2 ))= [Ostrogradski] =−(x/(24(x^3 −8)))−(1/(12))∫(dx/(x^3 −8))= =−(x/(24(x^3 −8)))−(1/(144))∫(dx/(x−2))+(1/(144))∫((x+4)/(x^2 +2x+4))dx= =−(x/(24(x^3 −8)))−((ln ∣x−2∣)/(144))+(1/(288))∫((2x+2)/(x^2 +2x+4))dx+(1/(48))∫(dx/(x^2 +2x+4))= =−(x/(24(x^3 −8)))−((ln ∣x−2∣)/(144))+((ln (x^2 +2x+4))/(288))+(((√3)arctan ((x+1)/( (√3))))/(144))+C
$$\int\frac{{dx}}{\left({x}^{\mathrm{3}} −\mathrm{8}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=−\frac{{x}}{\mathrm{24}\left({x}^{\mathrm{3}} −\mathrm{8}\right)}−\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{dx}}{{x}^{\mathrm{3}} −\mathrm{8}}= \\ $$$$=−\frac{{x}}{\mathrm{24}\left({x}^{\mathrm{3}} −\mathrm{8}\right)}−\frac{\mathrm{1}}{\mathrm{144}}\int\frac{{dx}}{{x}−\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{144}}\int\frac{{x}+\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}{dx}= \\ $$$$=−\frac{{x}}{\mathrm{24}\left({x}^{\mathrm{3}} −\mathrm{8}\right)}−\frac{\mathrm{ln}\:\mid{x}−\mathrm{2}\mid}{\mathrm{144}}+\frac{\mathrm{1}}{\mathrm{288}}\int\frac{\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}{dx}+\frac{\mathrm{1}}{\mathrm{48}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}= \\ $$$$=−\frac{{x}}{\mathrm{24}\left({x}^{\mathrm{3}} −\mathrm{8}\right)}−\frac{\mathrm{ln}\:\mid{x}−\mathrm{2}\mid}{\mathrm{144}}+\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)}{\mathrm{288}}+\frac{\sqrt{\mathrm{3}}\mathrm{arctan}\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{144}}+{C} \\ $$
Commented by JulioCesar last updated on 02/Apr/21
I don't understand sir

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