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Question Number 33590 by abdo imad last updated on 19/Apr/18

$$let\alpha >1calculatef\left(\alpha \right)={\int }_{\alpha }^{+\mathrm{\infty }}\frac{x^2-x+1}{{(x-1)}^2{(x+1)}^2}dx.$$

Commented byabdo imad last updated on 22/Apr/18

$$wehavef\left(\alpha \right)={\int }_{\alpha }^{+\mathrm{\infty }}\frac{x^2-2x+1+x}{{(x-1)}^2{(x+1)}^2}dx$$ $$={\int }_{\alpha }^{+\mathrm{\infty }}\frac{{(x-1)}^2+x}{{(x-1)}^2{(x+1)}^2}dx={\int }_{\alpha }^{+\mathrm{\infty }}\frac{dx}{{(x+1)}^2}+{\int }_{\alpha }^{+\mathrm{\infty }}\frac{x}{{(x-1)}^2{(x+1)}^2}dx$$ $$={\left[\frac{-1}{x+1}\right]}_{\alpha }^{+\mathrm{\infty }}+{\int }_{\alpha }^{+\mathrm{\infty }}\frac{xdx}{{(x-1)}^2{(x+1)}^2}=\frac{1}{\alpha +1}+IletfindI$$ $$F\left(x\right)=\frac{x}{{(x-1)}^2{(x+1)}^2}=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{c}{x+1}+\frac{d}{{(x+1)}^2}$$ $$b={lim}_{x\to 1}{(x-1)}^{2}F\left(x\right)=\frac{1}{4}$$ $$d={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=\frac{-1}{4}\Rightarrow$$ $$F\left(x\right)=\frac{a}{x-1}+\frac{1}{4{(x-1)}^2}+\frac{c}{x+1}+\frac{-1}{4{(x+1)}^2}$$ $$F\left(0\right)=0=-a+\frac{1}{4}+c-\frac{1}{4}\Rightarrow c=a\Rightarrow$$ $$F\left(x\right)=\frac{a}{x-1}+\frac{1}{4{(x-1)}^2}+\frac{a}{x+1}-\frac{1}{4{(x+1)}^2}$$ $$F\left(2\right)=\frac{2}{9}=a+\frac{1}{4}+\frac{a}{3}-\frac{1}{36}\Rightarrow 2=9a+\frac{9}{4}+3-\frac{1}{4}$$ $$\Rightarrow 2=9a+5\Rightarrow 9a=-3\Rightarrow a=-\frac{1}{3}\Rightarrow$$ $$F\left(x\right)=-\frac{1}{3(x-1)}+\frac{1}{4{(x-1)}^2}-\frac{1}{3(x+1)}-\frac{1}{4{(x+1)}^2}$$ $$I={\int }_{\alpha }^{+\mathrm{\infty }}(\frac{1}{3(1-x)}-\frac{1}{3(1+x)})dx+\frac{1}{4}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{dx}{{(x-1)}^2}$$ $$-\frac{1}{4}{\int }_{\alpha }^{+\mathrm{\infty }}\frac{dx}{{(x+1)}^2}={[\frac{1}{3}ln\mid \frac{1-x}{1+x}\mid ]}_{\alpha }^{+\mathrm{\infty }}-\frac{1}{4}{\left[\frac{1}{x-1}\right]}_{\alpha }^{+\mathrm{\infty }}$$ $$+\frac{1}{4}{\left[\frac{1}{x+1}\right]}_{\alpha }^{+\mathrm{\infty }}=\frac{1}{3}ln\mid \frac{1-\alpha }{1+\alpha }\mid +\frac{1}{4(\alpha -1)}-\frac{1}{4(\alpha +1)}\Rightarrow$$ $$f\left(\alpha \right)=\frac{3}{4(\alpha +1)}+\frac{1}{3}ln\mid \frac{1-\alpha }{1+\alpha }\mid +\frac{1}{4(\alpha -1)}.$$

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