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Question Number 95585 by turbo msup by abdo last updated on 26/May/20

$$calculate{\int }_2^{+\mathrm{\infty }}\frac{dx}{{(x-1)}^4{(x^2+x+1)}^2}$$$$$$

Answered by MJS last updated on 26/May/20

$$\int \frac{dx}{{(x-1)}^4{(x^2+x+1)}^2}=$$$$\left[\mathrm{Ostrogradski}\right]$$$$=-\frac{8x^4-11x^3-2x^2-x+9}{27{(x-1)}^4(x^2+x+1)}-\frac{2}{27}\int \frac{4x+5}{(x-1)(x^2+x+1)}dx$$$$$$$$-\frac{2}{27}\int \frac{4x+5}{(x-1)(x^2+x+1)}dx=$$$$=\frac{2}{27}\int \frac{3x+2}{x^2+x+1}dx-\frac{2}{9}\int \frac{dx}{x-1}=$$$$=\frac{2\sqrt{3}}{81}\arctan \frac{\sqrt{3}(2x+1)}{3}+\frac{1}{9}\ln (x^2+x+1)-\frac{2}{9}\ln (x-1)=$$$$=\frac{2\sqrt{3}}{81}\arctan \frac{\sqrt{3}(2x+1)}{3}+\frac{1}{9}\ln \frac{x^2+x+1}{{(x-1)}^2}+C$$$$$$$$\Rightarrow$$$$\underset{2}{\stackrel{+\mathrm{\infty }}{\int }}\frac{dx}{{(x-1)}^4{(x^2+x+1)}^2}=$$$$=\frac{13}{63}+\frac{\pi \sqrt{3}}{81}-\frac{\ln 7}{9}-\frac{2\sqrt{3}}{81}\arctan \frac{5\sqrt{3}}{3}$$

Commented by mathmax by abdo last updated on 27/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{mjs}$$

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