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Question Number 142185 by mohammad17 last updated on 27/May/21
∫(1/(x^4 +1))dx
$$\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$
Answered by MJS_new last updated on 27/May/21
I showed this before... x^4 +1=(x^2 −(√2)x+1)(x^2 +(√2)x+1) decompose and transform ∫(dx/(x^4 +1))=−((√2)/4)∫((x+(√2))/(x^2 −(√2)x+1))dx+((√2)/4)∫((x+(√2))/(x^2 +(√2)x+1))dx= =−((√2)/8)∫((2x−(√2))/(x^2 −(√2)x+1))dx+(1/4)∫(dx/(x^2 −(√2)x+1))+((√2)/8)∫((2x+(√2))/(x^2 +(√2)x+1))dx+(1/4)∫(dx/(x^2 +(√2)x+1))= =−((√2)/8)ln (x^2 −(√2)x+1) +((√2)/4)arctan ((√2)x−1) +((√2)/8)ln (x^2 +(√2)x+1) +((√2)/4)arctan ((√2)x+1) = =((√2)/8)ln ((x^2 +(√2)x+1)/(x^2 −(√2)x+1)) +((√2)/4)(arctan ((√2)x−1) +arctan ((√2)x+1)) +C
$$\mathrm{I}\:\mathrm{showed}\:\mathrm{this}\:\mathrm{before}… \\ $$$${x}^{\mathrm{4}} +\mathrm{1}=\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right) \\ $$$$\mathrm{decompose}\:\mathrm{and}\:\mathrm{transform} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{1}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)\:+{C} \\ $$
Answered by Ndala last updated on 29/May/21
I=(1/2)∫(2/(x^4 +1))dx=(1/2)∫(((x^2 +1)−(x^2 −1))/(x^4 +1))dx I=(1/2)∫((x^2 +1)/(x^4 +1))dx−(1/2)∫((x^2 −1)/(x^4 +1))dx I=(1/2)∫((1+x^(−2) )/(x^2 +x^(−2) +2−2))dx−(1/2)∫((1−x^(−2) )/(x^2 +x^(−2) +2−2))dx I=(1/2)∫(1/((x−x^(−1) )^2 +((√2))^2 ))d(x−x^(−1) )−(1/2)∫(1/((x+x^(−1) )^2 −((√2))^2 ))d(x+x^(−1) ) I=(1/(2(√2)))∙arctg(((x−x^(−1) )/( (√2))))−(1/(4(√2)))ln ∣((x+x^(−1) −(√2))/(x+x^(−1) +(√2)))∣+c I=(1/(2(√2)))∙arctg(((x^2 −1)/( x(√2))))−(1/(4(√2)))ln ∣((x^2 −x(√2)+1)/(x^2 +x(√2)+1))∣+c
$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{x}^{−\mathrm{2}} }{{x}^{\mathrm{2}} +{x}^{−\mathrm{2}} +\mathrm{2}−\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{x}^{−\mathrm{2}} }{{x}^{\mathrm{2}} +{x}^{−\mathrm{2}} +\mathrm{2}−\mathrm{2}}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}−{x}^{−\mathrm{1}} \right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\mathrm{d}\left({x}−{x}^{−\mathrm{1}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+{x}^{−\mathrm{1}} \overset{\mathrm{2}} {\right)}−\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\mathrm{d}\left(\mathrm{x}+{x}^{−\mathrm{1}} \right) \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\centerdot\mathrm{arctg}\left(\frac{\mathrm{x}−{x}^{−\mathrm{1}} }{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{x}+{x}^{−\mathrm{1}} −\sqrt{\mathrm{2}}}{{x}+{x}^{−\mathrm{1}} +\sqrt{\mathrm{2}}}\mid+\mathrm{c} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\centerdot\mathrm{arctg}\left(\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\:{x}\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{2}}+\mathrm{1}}\mid+\mathrm{c} \\ $$

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