Question Number 172526 by mnjuly1970 last updated on 28/Jun/22
$$ \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{\:{dx}}{\left(\mathrm{4}−\mathrm{2}{x}+{x}^{\:\mathrm{2}} \right)^{\:\mathrm{3}} }\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{64}}\:+\frac{\pi}{\mathrm{36}\sqrt{\mathrm{3}}} \\ $$
Answered by MJS_new last updated on 28/Jun/22
![∫(dx/((x^2 −2x+4)^3 ))= [Ostrogradski′s Method] =(((x−1)(x^2 −2x+6))/(24(x^2 −2x+4)^2 ))+(1/(24))∫(dx/(x^2 −2x+4))= =(((x−1)(x^2 −2x+6))/(24(x^2 −2x+4)^2 ))+((√3)/(72))arctan (((√3)(x−1))/3) +C ⇒ answer is true](https://www.tinkutara.com/question/Q172542.png)
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{6}\right)}{\mathrm{24}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{24}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{6}\right)}{\mathrm{24}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\right)^{\mathrm{2}} }+\frac{\sqrt{\mathrm{3}}}{\mathrm{72}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left({x}−\mathrm{1}\right)}{\mathrm{3}}\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{true} \\ $$