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Question-196459




Question Number 196459 by RoseAli last updated on 25/Aug/23
Answered by Frix last updated on 25/Aug/23
Use Ostrogradski′s Method to get  ∫((x^4 +x^2 +1)/((x^2 +1)^3 ))dx=((x(2x^2 +3))/(2(x^2 +1)))−(1/2)∫(dx/(x^2 +1))=  =((x(2x^2 +3))/(2(x^2 +1)))−((tan^(−1)  x)/2)+C  ⇒  For  (√2)−1≤x≤(√2)+1  the value of the integral is  2−(π/8)
$$\mathrm{Use}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{to}\:\mathrm{get} \\ $$$$\int\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx}=\frac{{x}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{{x}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{\mathrm{2}}+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{For} \\ $$$$\sqrt{\mathrm{2}}−\mathrm{1}\leqslant{x}\leqslant\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{is} \\ $$$$\mathrm{2}−\frac{\pi}{\mathrm{8}} \\ $$

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