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Question Number 82433 by mathmax by abdo last updated on 21/Feb/20

$$1)decomposeinsideC\left(x\right)andR\left(x\right)F=\frac{1}{{(x^2+x+1)}^2}$$$$2)calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^2+x+1)}^2}$$

Commented by mathmax by abdo last updated on 24/Feb/20

$$x^2+x+1=0\to \mathrm{\Delta }=1-4=-3\Rightarrow z_1=\frac{-1+i\sqrt{3}}{2}=e^{i\frac{2\pi }{3}}$$$$z_2=e^{-\frac{i2\pi }{3}}\Rightarrow F=\frac{1}{{(x-e^{\frac{i2\pi }{3}})}^2{(x-e^{-\frac{i2\pi }{3}})}^2}=\frac{a}{x-e^{\frac{i2\pi }{3}}}+\frac{b}{{(x-e^{\frac{i2\pi }{3}})}^2}$$$$+\frac{c}{(x-e^{-i\frac{2\pi }{3}})}+\frac{d}{{(x-e^{-i\frac{2\pi }{3}})}^2}$$$$b=\frac{1}{{\left(2isin\left(\frac{2\pi }{3}\right)\right)}^2}=-\frac{1}{4i{\left(\frac{\sqrt{3}}{2}\right)}^2}=-\frac{1}{3i}$$$$d=\frac{1}{{\left(2isin\left(\frac{2\pi }{3}\right)\right)}^2}=-\frac{1}{3i}\Rightarrow F\left(x\right)=\frac{a}{(x-e^{\frac{i2\pi }{3}})}-\frac{1}{3i{(x-e^{\frac{i2\pi }{3}})}^2}$$$$+\frac{c}{x-e^{-\frac{i2\pi }{3}}}-\frac{1}{3i{(x-e^{-\frac{i2\pi }{3}})}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a$$$$F\left(0\right)=1=-{ae}^{-\frac{i2\pi }{3}}-\frac{1}{3i}{e}^{-\frac{i4\pi }{3}}+{ae}^{\frac{i2\pi }{3}}-\frac{1}{3i}{e}^{\frac{i4\pi }{3}}$$$$=2iasin\left(\frac{2\pi }{3}\right)-\frac{1}{3i}\left(2cos\left(\frac{4\pi }{3}\right)\right)=2ia\times \frac{\sqrt{3}}{2}-\frac{2}{3i}(-\frac{1}{2})$$$$=ai\sqrt{3}+\frac{1}{3i}=ai\sqrt{3}-\frac{i}{3}=1\Rightarrow ai\sqrt{3}=1+\frac{i}{3}\Rightarrow$$$$a=\frac{1}{i\sqrt{3}}(1+\frac{i}{3})=\frac{1}{i\sqrt{3}}+\frac{1}{3\sqrt{3}}=\frac{1}{3\sqrt{3}}-\frac{i}{\sqrt{3}}andc=-\frac{1}{3\sqrt{3}}+\frac{i}{\sqrt{3}}\Rightarrow$$$$F\left(x\right)=(\frac{1}{3\sqrt{3}}-\frac{i}{\sqrt{3}})\times \frac{1}{x-e^{\frac{i2\pi }{3}}}+\frac{i}{3{(x-e^{\frac{i2\pi }{3}})}^2}-(\frac{1}{3\sqrt{3}}-\frac{i}{\sqrt{3}})\times \frac{1}{x-e^{-\frac{i2\pi }{3}}}$$$$+\frac{i}{3{(x-e^{-i\frac{2\pi }{3}})}^2}$$$$$$

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