Question Number 211696 by mnjuly1970 last updated on 16/Sep/24
$$ \\ $$$$\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\:\:\mathrm{1}}{\left(\:\mathrm{2}\:+\mathrm{2}{x}\:+\:{x}^{\mathrm{2}} \:\right)^{\mathrm{3}} }\:{dx}=\:? \\ $$$$\:\:\:\:\:\:\underbrace{\underset{\:\:\:\:\overset{\mathrm{Improper}\:\mathrm{integral}\:} {\:}\:\:\:\:\:} {\:}} \\ $$$$\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$
Answered by Ghisom last updated on 16/Sep/24
![∫(dx/((x^2 +2x+2)^3 ))= [Ostrogradski′s Method] =(((x+1)(3x^2 +6x+8))/(8(x^2 +2x+2)^2 ))+(3/8)∫(dx/(x^2 +2x+2))= =(((x+1)(3x^2 +6x+8))/(8(x^2 +2x+2)^2 ))+(3/8)arctan (x+1) +C ⇒ ∫_0 ^1 (dx/((x^2 +2x+2)^3 ))=−(2/(25))+arctan (1/3)](https://www.tinkutara.com/question/Q211697.png)
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{\left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)}{\mathrm{8}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{8}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}= \\ $$$$=\frac{\left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)}{\mathrm{8}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{8}}\mathrm{arctan}\:\left({x}+\mathrm{1}\right)\:+{C} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{3}} }=−\frac{\mathrm{2}}{\mathrm{25}}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$