Question Number 95585 by turbo msup by abdo last updated on 26/May/20
$${calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by MJS last updated on 26/May/20
![∫(dx/((x−1)^4 (x^2 +x+1)^2 ))= [Ostrogradski] =−((8x^4 −11x^3 −2x^2 −x+9)/(27(x−1)^4 (x^2 +x+1)))−(2/(27))∫((4x+5)/((x−1)(x^2 +x+1)))dx −(2/(27))∫((4x+5)/((x−1)(x^2 +x+1)))dx= =(2/(27))∫((3x+2)/(x^2 +x+1))dx−(2/9)∫(dx/(x−1))= =((2(√3))/(81))arctan (((√3)(2x+1))/3) +(1/9)ln (x^2 +x+1) −(2/9)ln (x−1) = =((2(√3))/(81))arctan (((√3)(2x+1))/3) +(1/9)ln ((x^2 +x+1)/((x−1)^2 )) +C ⇒ ∫_2 ^(+∞) (dx/((x−1)^4 (x^2 +x+1)^2 ))= =((13)/(63))+((π(√3))/(81))−((ln 7)/9)−((2(√3))/(81))arctan ((5(√3))/3)](Q95621.png)
$$\int\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=−\frac{\mathrm{8}{x}^{\mathrm{4}} −\mathrm{11}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{9}}{\mathrm{27}\left({x}−\mathrm{1}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{27}}\int\frac{\mathrm{4}{x}+\mathrm{5}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{dx} \\ $$$$ \\ $$$$−\frac{\mathrm{2}}{\mathrm{27}}\int\frac{\mathrm{4}{x}+\mathrm{5}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{27}}\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{\mathrm{2}}{\mathrm{9}}\int\frac{{dx}}{{x}−\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{81}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:−\frac{\mathrm{2}}{\mathrm{9}}\mathrm{ln}\:\left({x}−\mathrm{1}\right)\:= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{81}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+{C} \\ $$$$ \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{2}} {\overset{+\infty} {\int}}\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{13}}{\mathrm{63}}+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{81}}−\frac{\mathrm{ln}\:\mathrm{7}}{\mathrm{9}}−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{81}}\mathrm{arctan}\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 27/May/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mjs} \\ $$