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Question Number 99239 by abdomathmax last updated on 19/Jun/20

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

Answered by MJS last updated on 19/Jun/20

$$\mathrm{Ostrogradski}\mathrm{gives}$$$$-\frac{2x^2-1}{2x^2(x^2-1)}-2\int \frac{dx}{x(x^2-1)}=$$$$=-\frac{2x^2-1}{2x^2(x^2-1)}+\ln \mid \frac{x^2}{x^2-1}\mid +C$$$$\Rightarrow$$$$\underset{2}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x^3(x^2-1)}=\frac{7}{24}+\ln \frac{3}{4}$$

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

$$\mathrm{I}={\int }_2^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^3{({\mathrm{x}}^2-1)}^2}\Rightarrow \mathrm{I}={\int }_2^{\mathrm{\infty }}\frac{\mathrm{dx}}{{\left(\frac{\mathrm{x}}{\mathrm{x}-1}\right)}^3{(\mathrm{x}-1)}^5{(\mathrm{x}+1)}^2}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{\mathrm{x}}{\mathrm{x}-1}=\mathrm{t}\Rightarrow \mathrm{x}=\mathrm{tx}-\mathrm{t}\Rightarrow (1-\mathrm{t})\mathrm{x}=-\mathrm{t}\Rightarrow \mathrm{x}=\frac{\mathrm{t}}{\mathrm{t}-1}$$$$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}={(1+\frac{1}{\mathrm{t}-1})}^{{}^{\prime }}=-\frac{1}{{(\mathrm{t}-1)}^2}\mathrm{and}\mathrm{x}-1=\frac{\mathrm{t}}{\mathrm{t}-1}-1=\frac{\mathrm{t}-\mathrm{t}+1}{\mathrm{t}-1}=\frac{1}{\mathrm{t}-1}$$$$\mathrm{x}+1=\frac{\mathrm{t}}{\mathrm{t}-1}+1=\frac{\mathrm{t}+\mathrm{t}-1}{\mathrm{t}-1}=\frac{2\mathrm{t}-1}{\mathrm{t}-1}\Rightarrow$$$$\mathrm{I}=-{\int }_1^2\frac{-\mathrm{dt}}{{(\mathrm{t}-1)}^2{\mathrm{t}}^3{\left(\frac{1}{\mathrm{t}-1}\right)}^5{\left(\frac{2\mathrm{t}-1}{\mathrm{t}-1}\right)}^2}={\int }_1^2\frac{{(\mathrm{t}-1)}^7}{{(\mathrm{t}-1)}^2{\mathrm{t}}^3{(2\mathrm{t}-1)}^2}\mathrm{dt}$$$$={\int }_1^2\frac{{(\mathrm{t}-1)}^5}{{\mathrm{t}}^3{(2\mathrm{t}-1)}^2}\mathrm{dt}={\int }_1^2\frac{\sum _{\mathrm{k}=0}^5{\mathrm{C}}_5^{\mathrm{k}}{\mathrm{t}}^{\mathrm{k}}{(-1)}^{5-\mathrm{k}}}{{\mathrm{t}}^3{(2\mathrm{t}-1)}^2}\mathrm{dt}$$$$=-\sum _{\mathrm{k}=0}^5{\mathrm{C}}_5^{\mathrm{k}}{(-1)}^{\mathrm{k}}{\int }_1^2\frac{{\mathrm{t}}^{\mathrm{k}}}{{\mathrm{t}}^3{(2\mathrm{t}-1)}^2}\mathrm{dt}\mathrm{after}\mathrm{we}\mathrm{decompose}\mathrm{F}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^{\mathrm{k}}}{{\mathrm{t}}^3{(2\mathrm{t}-1)}^2}$$$$...\mathrm{be}\mathrm{continued}...$$

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