Question Number 98179 by abdomathmax last updated on 12/Jun/20
$$\mathrm{calculate}\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$
Answered by MJS last updated on 12/Jun/20
$$\mathrm{I}\:\mathrm{love}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}… \\ $$$$\mathrm{it}\:\mathrm{leads}\:\mathrm{to} \\ $$$$−\frac{\mathrm{15}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{8}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{15}}{\mathrm{8}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=−\frac{\mathrm{15}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{8}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{15}}{\mathrm{16}}\mathrm{ln}\:\mid\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\mid\:+{C} \\ $$$$\underset{\mathrm{2}} {\overset{+\infty} {\int}}\frac{{dx}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{37}}{\mathrm{36}}−\frac{\mathrm{15}}{\mathrm{16}}\mathrm{ln}\:\mathrm{3} \\ $$
Commented by mathmax by abdo last updated on 12/Jun/20
$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{mjs} \\ $$
Answered by mathmax by abdo last updated on 12/Jun/20
$$\mathrm{sorry}\:\mathrm{the}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{calculate}\:\:\int_{\mathrm{2}} ^{+\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$