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Question Number 99239 by abdomathmax last updated on 19/Jun/20
calculate ∫_2 ^(+∞) (dx/(x^3 (x^2 −1)^2 ))
$$\mathrm{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by MJS last updated on 19/Jun/20
Ostrogradski gives −((2x^2 −1)/(2x^2 (x^2 −1)))−2∫(dx/(x(x^2 −1)))= =−((2x^2 −1)/(2x^2 (x^2 −1)))+ln ∣(x^2 /(x^2 −1))∣ +C ⇒ ∫_2 ^∞ (dx/(x^3 (x^2 −1)))=(7/(24))+ln (3/4)
$$\mathrm{Ostrogradski}\:\mathrm{gives} \\ $$$$−\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}−\mathrm{2}\int\frac{{dx}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}= \\ $$$$=−\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}+\mathrm{ln}\:\mid\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\mid\:+{C} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{2}} {\overset{\infty} {\int}}\frac{{dx}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{7}}{\mathrm{24}}+\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by Ar Brandon last updated on 20/Jun/20
Hello Mr MJS,�� greetings to you Sir.
Commented by MJS last updated on 20/Jun/20
also back to you!
Answered by mathmax by abdo last updated on 21/Jun/20
I =∫_2 ^(+∞) (dx/(x^3 (x^2 −1)^2 )) ⇒ I =∫_2 ^∞ (dx/(((x/(x−1)))^3 (x−1)^5 (x+1)^2 )) we do the changement (x/(x−1)) =t ⇒x =tx−t ⇒(1−t)x =−t ⇒x =(t/(t−1)) ⇒(dx/dt) =(1+(1/(t−1)))^′ =−(1/((t−1)^2 )) and x−1 =(t/(t−1))−1 =((t−t+1)/(t−1)) =(1/(t−1)) x+1 =(t/(t−1)) +1 =((t+t−1)/(t−1)) =((2t−1)/(t−1)) ⇒ I =−∫_1 ^2 ((−dt)/((t−1)^2 t^3 ((1/(t−1)))^5 (((2t−1)/(t−1)))^2 )) =∫_1 ^2 (((t−1)^7 )/((t−1)^2 t^3 (2t−1)^2 ))dt =∫_1 ^2 (((t−1)^5 )/(t^3 (2t−1)^2 ))dt =∫_1 ^2 ((Σ_(k=0) ^5 C_5 ^k t^k (−1)^(5−k) )/(t^3 (2t−1)^2 ))dt =−Σ_(k=0) ^5 C_5 ^k (−1)^k ∫_1 ^2 (t^k /(t^3 (2t−1)^2 ))dt after we decompose F(t) =(t^k /(t^3 (2t−1)^2 )) ...be continued...
$$\mathrm{I}\:=\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:\mathrm{I}\:=\int_{\mathrm{2}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{3}} \:\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\:=\mathrm{t}\:\Rightarrow\mathrm{x}\:=\mathrm{tx}−\mathrm{t}\:\Rightarrow\left(\mathrm{1}−\mathrm{t}\right)\mathrm{x}\:=−\mathrm{t}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{t}}{\mathrm{t}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}\:=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}\right)^{'} \:=−\frac{\mathrm{1}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{and}\:\mathrm{x}−\mathrm{1}\:=\frac{\mathrm{t}}{\mathrm{t}−\mathrm{1}}−\mathrm{1}\:=\frac{\mathrm{t}−\mathrm{t}+\mathrm{1}}{\mathrm{t}−\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}} \\ $$$$\mathrm{x}+\mathrm{1}\:=\frac{\mathrm{t}}{\mathrm{t}−\mathrm{1}}\:+\mathrm{1}\:=\frac{\mathrm{t}+\mathrm{t}−\mathrm{1}}{\mathrm{t}−\mathrm{1}}\:=\frac{\mathrm{2t}−\mathrm{1}}{\mathrm{t}−\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{−\mathrm{dt}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}\right)^{\mathrm{5}} \left(\frac{\mathrm{2t}−\mathrm{1}}{\mathrm{t}−\mathrm{1}}\right)^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{7}} }{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{3}} \left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{5}} }{\mathrm{t}^{\mathrm{3}} \left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} \:\mathrm{t}^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{5}−\mathrm{k}} }{\mathrm{t}^{\mathrm{3}} \left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=−\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{C}_{\mathrm{5}} ^{\mathrm{k}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{t}^{\mathrm{k}} }{\mathrm{t}^{\mathrm{3}} \left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}\:\:\mathrm{after}\:\mathrm{we}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{t}^{\mathrm{k}} }{\mathrm{t}^{\mathrm{3}} \left(\mathrm{2t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$

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