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Question Number 154780 by talminator2856791 last updated on 21/Sep/21

$$$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{x^4+x^3+x^2+1}dx$$$$$$

Answered by phanphuoc last updated on 21/Sep/21

$$youcanfindabcdst$$$$x^4+x^3+x^2+1=(x^2+ax+b)(x^2+cx+d)$$

Answered by MJS_new last updated on 21/Sep/21

$$(x-1)(x^4+x^3+x^2+x+1)=x^5-1$$$$\Rightarrow$$$$x^4+x^3+x^2+x+1=(x^2+\frac{1-\sqrt{5}}{2}x+1)(x^2+\frac{1+\sqrt{5}}{2}x+1)$$$$\int \frac{dx}{x^4+x^3+x^2+x+1}=$$$$=-\frac{\sqrt{5}}{10}\int \frac{2x+1-\sqrt{5}}{x^2+\frac{1-\sqrt{5}}{2}x+1}dx+\frac{\sqrt{5}}{10}\int \frac{2x+1+\sqrt{5}}{x^2+\frac{1+\sqrt{5}}{2}x+1}dx$$$$\mathrm{now}\mathrm{use}\mathrm{formula}$$

Commented by Ar Brandon last updated on 21/Sep/21

$$\mathrm{Sir}\mathrm{I}\mathrm{thought}\mathrm{of}\mathrm{doing}\mathrm{same}\mathrm{before}\mathrm{realising}\mathrm{that}\mathrm{it}\mathrm{was}$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{x^4+x^3+x^2+1}\mathrm{instead}\mathrm{of}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{x^4+x^3+x^2+x+1}$$

Commented by MJS_new last updated on 21/Sep/21

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}...\mathrm{we}\mathrm{see}\mathrm{what}\mathrm{we}\mathrm{like}\mathrm{to}\mathrm{see}...$$

Commented by Rasheed.Sindhi last updated on 21/Sep/21

$$\mathcal{G}ood\mathcal{S}aying:\mathrm{we}\mathrm{see}\mathrm{what}\mathrm{we}\mathrm{like}\mathrm{to}\mathrm{see}!$$

Answered by mindispower last updated on 22/Sep/21

$$=2i\pi \sum _{w\in \{1+w^2+w^3+w^4=0\}}\frac{1}{4w^3+3w^2+2w}$$

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