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Question Number 129921 by stelor last updated on 20/Jan/21

∫(1/((1−u^2 )^2 u^5 ))du=?? please...

$$\int\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}^{\mathrm{5}} }\mathrm{du}=?? \\ $$$$\mathrm{please}... \\ $$

Answered by Ar Brandon last updated on 20/Jan/21

u=cosθ I=−∫((sinθdθ)/(sin^4 θcos^5 θ))=−∫(dθ/(sin^3 θcos^5 θ)) see Q129907 for similar case

$$\mathrm{u}=\mathrm{cos}\theta \\ $$$$\mathcal{I}=−\int\frac{\mathrm{sin}\theta\mathrm{d}\theta}{\mathrm{sin}^{\mathrm{4}} \theta\mathrm{cos}^{\mathrm{5}} \theta}=−\int\frac{\mathrm{d}\theta}{\mathrm{sin}^{\mathrm{3}} \theta\mathrm{cos}^{\mathrm{5}} \theta} \\ $$$${see}\:{Q}\mathrm{129907}\:\mathrm{for}\:\mathrm{similar}\:\mathrm{case} \\ $$

Answered by Ar Brandon last updated on 20/Jan/21

I=∫(du/((1−u^2 )^2 u^5 )) f(u)=(1/((1−u^2 )^2 u^5 ))=(((1−u^2 )+u^2 )/((1−u^2 )^2 u^5 ))=(1/((1−u^2 )u^5 ))+(1/((1−u^2 )^2 u^3 )) =(((1−u^2 )+u^2 )/((1−u^2 )u^5 ))+(((1−u^2 )+u^2 )/((1−u^2 )^2 u^3 )) =(1/u^5 )+(1/((1−u^2 )u^3 ))+(1/((1−u^2 )u^3 ))+(1/((1−u^2 )^2 u)) =(1/u^5 )+(((1−u^2 )+u^2 )/((1−u^2 )u^3 ))+(((1−u^2 )+u^2 )/((1−u^2 )u^3 ))+(((1−u^2 )+u^2 )/((1−u^2 )^2 u)) =(1/u^5 )+(1/u^3 )+(1/((1−u^2 )u))+(1/u^3 )+(1/((1−u^2 )u))+(1/((1−u^2 )u))+(u/((1−u^2 )^2 )) =(1/u^5 )+(2/u^3 )+3((1/u)+(u/(1−u^2 )))+(u/((1−u^2 )^2 )) ∫f(u)du=−(1/(4u^4 ))−(1/u^2 )+3ln∣u∣−((3ln(1−u^2 ))/2)+(1/(2(1−u^2 )))+C

$$\mathcal{I}=\int\frac{\mathrm{du}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}^{\mathrm{5}} } \\ $$$$\mathrm{f}\left(\mathrm{u}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}^{\mathrm{5}} }=\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)+\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}^{\mathrm{5}} }=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}^{\mathrm{5}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)+\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}^{\mathrm{5}} }+\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)+\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{5}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{5}} }+\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)+\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}^{\mathrm{3}} }+\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)+\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}^{\mathrm{3}} }+\frac{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)+\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{u}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}}+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}}+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}}+\frac{\mathrm{u}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{5}} }+\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{3}} }+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{u}}+\frac{\mathrm{u}}{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\right)+\frac{\mathrm{u}}{\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\int\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}=−\frac{\mathrm{1}}{\mathrm{4u}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }+\mathrm{3ln}\mid\mathrm{u}\mid−\frac{\mathrm{3ln}\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)}+\mathcal{C} \\ $$

Commented by stelor last updated on 21/Jan/21

merci ...thank...

$${merc}\mathrm{i}\:\:...\mathrm{thank}... \\ $$

Answered by MJS_new last updated on 20/Jan/21

∫(du/(u^5 (1−u^2 )^2 ))= [Ostrogradski′s Method] =−((6u^4 −3u^2 −1)/(4u^4 (u^2 −1)))−3∫(du/(u(u^2 −1)))= =−((6u^4 −3u^2 −1)/(4u^4 (u^2 −1)))−(3/2)∫(du/(u−1))−(3/2)∫(du/(u+1))+3∫(du/u)= =−((6u^4 −3u^2 −1)/(4u^4 (u^2 −1)))−(3/2)ln ∣u−1∣ −(3/2)ln ∣u+1∣ +3ln ∣u∣ = =−((6u^4 −3u^2 −1)/(4u^4 (u^2 −1)))+(3/2)ln (u^2 /(∣u^2 −1∣)) +C

$$\int\frac{{du}}{{u}^{\mathrm{5}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=−\frac{\mathrm{6}{u}^{\mathrm{4}} −\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{u}^{\mathrm{4}} \left({u}^{\mathrm{2}} −\mathrm{1}\right)}−\mathrm{3}\int\frac{{du}}{{u}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}= \\ $$$$=−\frac{\mathrm{6}{u}^{\mathrm{4}} −\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{u}^{\mathrm{4}} \left({u}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{du}}{{u}−\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{du}}{{u}+\mathrm{1}}+\mathrm{3}\int\frac{{du}}{{u}}= \\ $$$$=−\frac{\mathrm{6}{u}^{\mathrm{4}} −\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{u}^{\mathrm{4}} \left({u}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mid{u}−\mathrm{1}\mid\:−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mid{u}+\mathrm{1}\mid\:+\mathrm{3ln}\:\mid{u}\mid\:= \\ $$$$=−\frac{\mathrm{6}{u}^{\mathrm{4}} −\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}{u}^{\mathrm{4}} \left({u}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\frac{{u}^{\mathrm{2}} }{\mid{u}^{\mathrm{2}} −\mathrm{1}\mid}\:+{C} \\ $$

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