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Question Number 76360 by mathmax by abdo last updated on 26/Dec/19

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

Commented by abdomathmax last updated on 29/Dec/19

$$letdecomposeF\left(x\right)=\frac{1}{x^3(x^2-2x+3)}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3}+\frac{dx+e}{x^2-2x+3}$$$$c=\frac{1}{3}{lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+d\Rightarrow d=-a\Rightarrow$$$$f\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{1}{3x^3}+\frac{-ax+e}{x^2-2x+3}$$$$f\left(x\right)-\frac{1}{3x^3}=\frac{a}{x}+\frac{b}{x^2}+\frac{-ax+e}{x^2-2x+3}\Rightarrow$$$$\frac{1}{x^3(x^2-2x+3)}-\frac{1}{3x^3}=...\Rightarrow$$$$\frac{3x^3-x^5+2x^4-3x^3}{3x^6(x^2-2x+3)}=...\Rightarrow$$$$\frac{-x+2}{3x^2(x^2-2x+3)}=\frac{a}{x}+\frac{b}{x^2}+\frac{-ax+e}{x^2-2x+3}$$$$b=\frac{2}{9}\Rightarrow f\left(x\right)-\frac{1}{3x^3}=\frac{a}{x}+\frac{2}{9x^2}+\frac{-ax+e}{x^2-2x+3}\Rightarrow$$$$f\left(x\right)=\frac{a}{x}+\frac{2}{9x^2}+\frac{1}{3x^3}+\frac{-ax+e}{x^2-2x+3}$$$$f\left(1\right)=\frac{1}{3}=a+\frac{2}{9}+\frac{1}{3}+\frac{-a+e}{2}\Rightarrow$$$$\frac{a}{2}+\frac{2}{9}+\frac{e}{2}=0\Rightarrow \frac{a+e}{2}=-\frac{2}{9}\Rightarrow a+e=-\frac{4}{9}$$$$f(-1)=-\frac{1}{6}=-a+\frac{2}{9}-\frac{1}{3}+\frac{a+e}{6}\Rightarrow$$$$-\frac{5}{6}a+\frac{e}{6}-\frac{1}{9}=-\frac{1}{6}\Rightarrow -5a+e-\frac{2}{3}=-1\Rightarrow$$$$-5a+e=-\frac{1}{3}\Rightarrow 5a-e=\frac{1}{3}\Rightarrow 5a-(-a-\frac{4}{9})=\frac{1}{3}$$$$\Rightarrow 6a+\frac{4}{9}=\frac{1}{3}\Rightarrow 6a=\frac{1}{3}-\frac{4}{9}=-\frac{1}{9}\Rightarrow a=-\frac{1}{54}$$$$e=-\frac{4}{9}-a=\frac{1}{54}-\frac{4}{9}=\frac{-23}{54}\Rightarrow$$$$f\left(x\right)=-\frac{1}{54x}+\frac{2}{9x^2}+\frac{1}{3x^3}+\frac{\frac{x}{54}-\frac{23}{54}}{x^2-2x+3}\Rightarrow$$$$\int f\left(x\right)dx=-\frac{1}{54}ln\mid x\mid -\frac{2}{9x}+\frac{1}{3}\times \frac{1}{-3+1}{x}^{-3+1}$$$$+\frac{1}{54}\int \frac{x-23}{x^2-2x+3}dx....becontinued...$$$$$$

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