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1-1-x-4-dx-

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Question Number 50796 by lihao12345 last updated on 20/Dec/18
∫1/(1+x^4 )dx=
$$\int\mathrm{1}/\left(\mathrm{1}+{x}^{\mathrm{4}} \right){dx}= \\ $$
Commented by Abdo msup. last updated on 20/Dec/18
this integral is solved i give the key we decompose F(x)=(1/(x^4 +1)) F(x)=(1/((x^2 +1)^2 −2x)) =(1/((x^2 −(√2)x +1)(x^2 +(√2)x +1))) =((ax+b)/(x^2 −(√2)x +1)) +((cx+d)/(x^2 +(√2)x +1)) F(−x)=F(x) ⇒((−ax+b)/(x^2 +(√2)x +1)) +((−cx +d)/(x^2 −(√2)x +1))=F(x)⇒ a=−c and b=d ⇒ F(x)=((ax+b)/(x^2 −(√2)x +1)) −((ax−b)/(x^2 +(√2)x +1)) F(0) =1 =2b ⇒b=(1/2) F(1) =(1/2) =((a+b)/(2−(√2)))−((a−b)/(2+(√2))) =(((2+(√2))a+(2+(√2))b−(2−(√2))a+(2−(√2))b)/2) =((2(√2)a +4b)/2) ⇒2(√2)a +2 =1 ⇒2(√2)a=−1 ⇒ a =((−1)/(2(√2))) ⇒ F(x) = ((((−x)/(2(√2)))+(1/2))/(x^2 −(√2)x +1)) −((−(x/(2(√2)))−(1/2))/(x^2 +(√2)x +1)) =−(1/(2(√2))) ((x−(√2))/(x^2 −(√2)x +1)) +(1/(2(√2))) ((x+(√2))/(x^2 +(√2)x +1)) ⇒ ∫ F(x)dx =(1/(2(√2))) ∫ ((x+(√2))/(x^2 +(√2)x +1))dx −(1/(2(√2))) ∫ ((x−(√2))/(x^2 −(√2)x +1))dx =(1/(4(√2))) ∫ ((2x +(√2)+(√2))/(x^2 +(√2)x+1))dx−(1/(4(√2))) ∫ ((2x−(√2)−(√2))/(x^2 −(√2)x +1))dx =(1/(4(√2)))ln(x^2 +(√2)x +1)+(1/4) ∫ (dx/(x^2 +(√2)x +1)) −(1/(4(√2)))ln(x^2 +(√2)x +1) +(1/4) ∫ (dx/(x^2 −(√2)x +1)) =.....
$${this}\:{integral}\:{is}\:{solved}\:\:{i}\:{give}\:{the}\:{key} \\ $$$${we}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}}\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)} \\ $$$$=\frac{{ax}+{b}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$${F}\left(−{x}\right)={F}\left({x}\right)\:\Rightarrow\frac{−{ax}+{b}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:+\frac{−{cx}\:+{d}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}={F}\left({x}\right)\Rightarrow \\ $$$${a}=−{c}\:{and}\:{b}={d}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{ax}+{b}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:−\frac{{ax}−{b}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\mathrm{2}{b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{{a}+{b}}{\mathrm{2}−\sqrt{\mathrm{2}}}−\frac{{a}−{b}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:=\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){a}+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){b}−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){b}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}{a}\:+\mathrm{4}{b}}{\mathrm{2}}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}{a}\:+\mathrm{2}\:=\mathrm{1}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}{a}=−\mathrm{1}\:\Rightarrow \\ $$$${a}\:=\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{\frac{−{x}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:−\frac{−\frac{{x}}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\frac{{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx}\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\frac{{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\int\:\:\frac{\mathrm{2}{x}\:+\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\int\:\:\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$$=….. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Dec/18
∫(dx/(1+x^4 )) ∫((1/x^2 )/(x^2 +(1/x^2 )))dx (1/2)∫(((1+(1/x^2 ))−(1−(1/x^2 )))/(x^2 +(1/x^2 )))dx (1/2)∫((d(x−(1/x)))/((x−(1/x))^2 +2))−(1/2)∫((d(x+(1/x)))/((x+(1/x))^2 −2)) (1/2)×(1/( (√2)))tan^(−1) (((x−(1/x))/( (√2))))−(1/2)×(1/(2(√2)))ln(((x+(1/x)−(√2))/(x+(1/x)+(√2))))+c pls check...
$$\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$\int\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right)+{c} \\ $$$${pls}\:{check}… \\ $$

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