Question Number 135693 by liberty last updated on 15/Mar/21
$$\Omega\:=\:\int\:\frac{{x}−\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\:{dx}\: \\ $$
Answered by MJS_new last updated on 15/Mar/21
![∫((x−1)/((x−2)(x^2 −2x+2)^2 ))dx= [Ostrogradski] =(x/(4(x^2 −2x+2)))+(1/4)∫(x/((x−2)(x^2 −2x+2)))dx= =(x/(4(x^2 −2x+2)))+(1/4)∫(dx/(x−2))−(1/4)∫((x−1)/(x^2 −2x+2))dx= =(x/(4(x^2 −2x+2)))+(1/4)ln ∣x−2∣ −(1/8)ln (x^2 −2x+2) +C (=(x/(4(x^2 −2x+2)))+(1/8)ln (((x−2)^2 )/(x^2 −2x+2)) +C)](https://www.tinkutara.com/question/Q135705.png)
$$\int\frac{{x}−\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{{x}}{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx}= \\ $$$$=\frac{{x}}{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}= \\ $$$$=\frac{{x}}{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{x}−\mathrm{2}\mid\:−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\:+{C} \\ $$$$\left(=\frac{{x}}{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}\:+{C}\right) \\ $$
Commented by liberty last updated on 15/Mar/21
$${nice}\:{sir} \\ $$