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

$$\mathrm{calculate}{\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2{(1-{\mathrm{x}}^2)}^3}$$

Answered by MJS last updated on 12/Jun/20

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

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

$$\mathrm{sorry}\mathrm{the}\mathrm{Q}\mathrm{is}\mathrm{calculate}{\int }_2^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2{(1-{\mathrm{x}}^2)}^3}$$

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