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Question Number 130029 by mathmax by abdo last updated on 22/Jan/21

$$1)\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{({\mathrm{x}}^2-4)}^3{({\mathrm{x}}^2+1)}^2}$$$$2)\mathrm{find}{\int }_3^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2-4)}^3{({\mathrm{x}}^2+1)}^2}$$

Commented by MJS_new last updated on 22/Jan/21

$$\mathrm{Ostrogradski}\mathrm{gives}$$$$F\left(x\right)=-\frac{x(17x^4-331x^2+1252)}{16000{(x^2-4)}^2(x^2+1)}-\frac{1}{16000}\int \frac{17x^2-687}{(x-2)(x+2)(x^2+1)}dx=$$$$=-\frac{x(17x^4-331x^2+1252)}{16000{(x^2-4)}^2(x^2+1)}+\frac{619}{320000}\int \frac{1}{x+2}-\frac{1}{x-2}dx-\frac{11}{1250}\int \frac{dx}{x^2+1}=$$$$=-\frac{x(17x^4-331x^2+1252)}{16000{(x^2-4)}^2(x^2+1)}+\frac{619}{320000}\ln \mid \frac{x-2}{x+2}\mid -\frac{11}{1250}\arctan x+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{619}{320000}\ln 5-\frac{11}{1250}\arctan \frac{1}{3}-\frac{21}{80000}$$

Commented by liberty last updated on 22/Jan/21

$$\mathrm{your}\mathrm{favorite}\mathrm{method}\mathrm{sir}\mathrm{hahaha}$$

Answered by Ar Brandon last updated on 22/Jan/21

$$\mathrm{f}(\mathrm{a},\mathrm{b})$$$$\mathrm{f}(\mathrm{a},\mathrm{b})=\frac{1}{({\mathrm{x}}^2-{\mathrm{a}}^2)({\mathrm{x}}^2+{\mathrm{b}}^2)}=\frac{\alpha }{\mathrm{x}-\mathrm{a}}+\frac{\beta }{\mathrm{x}+\mathrm{a}}+\frac{\lambda \mathrm{x}+\mu }{{\mathrm{x}}^2+{\mathrm{b}}^2}$$$$=\frac{\alpha (\mathrm{x}+\mathrm{a})({\mathrm{x}}^2+{\mathrm{b}}^2)+\beta (\mathrm{x}-\mathrm{a})({\mathrm{x}}^2+{\mathrm{b}}^2)+(\lambda \mathrm{x}+\mu )({\mathrm{x}}^2-{\mathrm{a}}^2)}{({\mathrm{x}}^2-{\mathrm{a}}^2)({\mathrm{x}}^2+{\mathrm{b}}^2)}$$$$\alpha =\frac{1}{2\mathrm{a}({\mathrm{a}}^2+{\mathrm{b}}^2)},\beta =\frac{1}{-2\mathrm{a}({\mathrm{a}}^2+{\mathrm{b}}^2)}$$$$\alpha {\mathrm{ab}}^2-\beta {\mathrm{ab}}^2-\mu {\mathrm{a}}^2=1\Rightarrow \mu {\mathrm{a}}^2=\left(\frac{{\cancel{\mathrm{a}b}}^2}{2\cancel{\mathrm{a}}({\mathrm{a}}^2+{\mathrm{b}}^2)}\right)+\left(\frac{{\cancel{\mathrm{a}b}}^2}{2\cancel{\mathrm{a}}({\mathrm{a}}^2+{\mathrm{b}}^2)}\right)-1$$$$\Rightarrow \mu =-\frac{1}{({\mathrm{a}}^2+{\mathrm{b}}^2)},\alpha +\beta +\lambda =0\Rightarrow \lambda =0$$$$\mathrm{f}(\mathrm{a},\mathrm{b})=\frac{1}{2\mathrm{a}({\mathrm{a}}^2+{\mathrm{b}}^2)}(\frac{1}{\mathrm{x}-\mathrm{a}}-\frac{1}{\mathrm{x}+\mathrm{a}})-\frac{1}{({\mathrm{a}}^2+{\mathrm{b}}^2)}\cdot \frac{1}{{\mathrm{x}}^2+{\mathrm{b}}^2}$$$$\int \mathrm{f}(\mathrm{a},\mathrm{b})\mathrm{dx}=\frac{\ln \mid \mathrm{x}-\mathrm{a}\mid -\ln \mid \mathrm{x}+\mathrm{a}\mid }{2\mathrm{a}({\mathrm{a}}^2+{\mathrm{b}}^2)}-\frac{{\tan }^{-1}\left(\mathrm{b}/\mathrm{x}\right)}{\mathrm{b}({\mathrm{a}}^2+{\mathrm{b}}^2)}+\mathcal{C}$$$$\mathrm{F}\left(\mathrm{x}\right)=\int \{\frac{{\partial }^3\mathrm{f}(\mathrm{a},\mathrm{b})}{\partial {\mathrm{a}}^3}\cdot \frac{{\partial }^2\mathrm{f}(\mathrm{a},\mathrm{b})}{\partial {\mathrm{b}}^2}\}\mathrm{dx}$$

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