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Question Number 98179 by abdomathmax last updated on 12/Jun/20

calculate ∫_1 ^(+∞) (dx/(x^2 (1−x^2 )^3 ))

$$\mathrm{calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$

Answered by MJS last updated on 12/Jun/20

I love Ostrogradski′s Method... it leads to −((15x^4 −25x^2 +8)/(8x(x^2 −1)^2 ))−((15)/8)∫(dx/(x^2 −1))= =−((15x^4 −25x^2 +8)/(8x(x^2 −1)^2 ))+((15)/(16))ln ∣((x+1)/(x−1))∣ +C ∫_2 ^(+∞) (dx/(x^2 (1−x^2 )^3 ))=((37)/(36))−((15)/(16))ln 3

$$\mathrm{I}\:\mathrm{love}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}... \\ $$$$\mathrm{it}\:\mathrm{leads}\:\mathrm{to} \\ $$$$−\frac{\mathrm{15}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{8}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{15}}{\mathrm{8}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=−\frac{\mathrm{15}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{8}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{15}}{\mathrm{16}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\mid\:+{C} \\ $$$$\underset{\mathrm{2}} {\overset{+\infty} {\int}}\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{37}}{\mathrm{36}}−\frac{\mathrm{15}}{\mathrm{16}}\mathrm{ln}\:\mathrm{3} \\ $$

Commented by mathmax by abdo last updated on 12/Jun/20

thanks sir mjs

$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{mjs} \\ $$

Answered by mathmax by abdo last updated on 12/Jun/20

sorry the Q is calculate ∫_2 ^(+∞) (dx/(x^2 (1−x^2 )^3 ))

$$\mathrm{sorry}\:\mathrm{the}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{calculate}\:\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$

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