Question Number 212052 by Spillover last updated on 28/Sep/24
Answered by Ghisom last updated on 28/Sep/24
![∫(dx/((x+(1/x))^2 ))= [Ostrogradski′s M ethod] =−(x/(2(x^2 +1)))+(1/2)∫(dx/(x^2 +1))= =(1/2)arctan x −(x/(2(x^2 +1)))+C ∫_0 ^∞ (dx/((x+(1/x))^2 ))=(π/4)](https://www.tinkutara.com/question/Q212074.png)
$$\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{M}\:\mathrm{ethod}\right] \\ $$$$=−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\:{x}\:−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}+{C} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{4}} \\ $$