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Question Number 81433 by abdomathmax last updated on 13/Feb/20
calculate ∫_2 ^(+∞) ((2x+3)/((x−1)^3 (x^2 +x+1)^2 ))dx
$${calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{2}{x}+\mathrm{3}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Answered by MJS last updated on 13/Feb/20
∫((2x+3)/((x−1)^2 (x^2 +x+1)^2 ))= Q_1 (x)=Q_2 (x)=(x−1)(x^2 +x+1) P_1 (x)=−(2/3)x^2 −x P_2 (x)=−(2/3)x−2 =−((x(2x+3))/(3(x−1)(x^2 +x+1)))−(2/3)∫((x+3)/((x−1)(x^2 +x+1)))dx= ∫((x+3)/((x−1)(x^2 +x+1)))dx=(4/3)∫(dx/(x−1))−(1/3)∫((4x+5)/(x^2 +x+1))dx= =(4/3)ln ∣x−1∣ −(2/3)ln (x^2 +x+1) −((2(√3))/3)arctan (((√3)(2x+1))/3) = =(2/3)ln (((x−1)^2 )/(x^2 +x+1)) −((2(√3))/3)arctan (((√3)(2x+1))/3) =−((x(2x+3))/(3(x−1)(x^2 +x+1)))−(4/9)ln (((x−1)^2 )/(x^2 +x+1)) +((4(√3))/9)arctan (((√3)(2x+1))/3) +C ⇒ answer is ((2π(√3))/9)+(2/3)−(4/9)ln 7 −((4(√3))/9)arctan ((5(√3))/3)
$$\int\frac{\mathrm{2}{x}+\mathrm{3}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:{Q}_{\mathrm{1}} \left({x}\right)={Q}_{\mathrm{2}} \left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:{P}_{\mathrm{1}} \left({x}\right)=−\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{2}} −{x} \\ $$$$\:\:\:\:\:{P}_{\mathrm{2}} \left({x}\right)=−\frac{\mathrm{2}}{\mathrm{3}}{x}−\mathrm{2} \\ $$$$=−\frac{{x}\left(\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{3}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{x}+\mathrm{3}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\int\frac{{x}+\mathrm{3}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}{dx}=\frac{\mathrm{4}}{\mathrm{3}}\int\frac{{dx}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{4}{x}+\mathrm{5}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\mathrm{ln}\:\mid{x}−\mathrm{1}\mid\:−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:= \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\:\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$=−\frac{{x}\left(\mathrm{2}{x}+\mathrm{3}\right)}{\mathrm{3}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}}\:+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{9}}\mathrm{ln}\:\mathrm{7}\:−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
Commented by john santu last updated on 13/Feb/20
Ostrogradski method. i will learn this method
$$\mathrm{Ostrogradski}\:\mathrm{method}.\:\mathrm{i}\:\mathrm{will}\:\mathrm{learn} \\ $$$$\mathrm{this}\:\mathrm{method} \\ $$$$ \\ $$
Commented by abdomathmax last updated on 13/Feb/20
thank you sir mjs
$${thank}\:{you}\:{sir}\:{mjs} \\ $$

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