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Question-137290




Question Number 137290 by JulioCesar last updated on 31/Mar/21
Answered by EDWIN88 last updated on 01/Apr/21
Ostrogradski Integral
$$\mathrm{Ostrogradski}\:\mathrm{Integral}\: \\ $$
Answered by MJS_new last updated on 01/Apr/21
Ostrogradski gives  −(x^3 /(4(x^4 +1)))−(1/4)∫((x^2 −4)/(x^4 +1))dx  this Integral decomposed is  −((5(√2))/(32))∫((2x−(√2))/(x^2 −(√2)x+1))+(3/(16))∫(dx/(x^2 −(√2)x+1))+((5(√2))/(32))∫((2x+(√2))/(x^2 +(√2)x+1))+(3/(16))∫(dx/(x^2 +(√2)x+1))  now solve with obvious formulas
$$\mathrm{Ostrogradski}\:\mathrm{gives} \\ $$$$−\frac{{x}^{\mathrm{3}} }{\mathrm{4}\left({x}^{\mathrm{4}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$$$\mathrm{this}\:\mathrm{Integral}\:\mathrm{decomposed}\:\mathrm{is} \\ $$$$−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{32}}\int\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{16}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{32}}\int\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{16}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}} \\ $$$$\mathrm{now}\:\mathrm{solve}\:\mathrm{with}\:\mathrm{obvious}\:\mathrm{formulas} \\ $$

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