1-1-x-4-dx-

Question Number 44696 by manish09@gmail.com last updated on 03/Oct/18

$$\int\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{4}} }\boldsymbol{\mathrm{dx}}\:=\:? \\ $$

Commented by maxmathsup by imad last updated on 03/Oct/18

let decompose F(x)=(1/(1+x^4 )) F(x)=(1/((x^2 +1)^2 −2x^2 )) =(1/((x^2 +1−(√2)x)(x^2 +1 +(√2)x))) = ((ax+b)/((x^2 −x(√2) +1))) +((cx +d)/(x^2 +x(√2) +1)) we have F(−x)=F(x) ⇒ ((−ax+b)/(x^2 +x(√2)+1)) +((−cx +d)/(x^2 −x(√2)+1)) =F(x) ⇒c=−a and b=d ⇒ F(x)=((ax+b)/(x^2 −x(√2)+1)) +((−ax +b)/(x^2 +x(√2) +1)) F(0) =1 =2b ⇒b =(1/2) F(1) =(1/2) = ((a+b)/(2−(√2))) +((−a +b)/(2+(√2))) ⇒(1/2) = (((2+(√2))(a+b)+(2−(√2))(−a+b))/2) ⇒ (2+(√2)−2+(√2))a +(2+(√2)+2−(√2))b =1 ⇒ 2(√2)a +4b =1 ⇒2(√2)a +2=1 ⇒a=−(1/(2(√2))) ⇒ F(x) =(((1/(2(√2)))x +(1/2))/(x^2 +x(√2) +1)) + ((−(1/(2(√2) ))x +(1/2))/(x^2 −x(√2)+1)) =(1/(2(√2))){ ((x+(√2))/(x^2 +x(√2) +1)) +((−x+(√2))/(x^2 −x(√2) +1))} ⇒ ∫ F(x)dx =(1/(4(√2)))∫{ ((2x+2(√2))/(x^2 +x(√2) +1)) −((2x−2(√2))/(x^2 −x(√2)+1))}dx = (1/(4(√2))) ∫ ((2x+(√2))/(x^2 +x(√2) +1))dx +(1/2) ∫ (dx/(x^2 +x(√2)+1)) −(1/(4(√2)))∫ ((2x−(√2))/(x^2 −x(√2) +1))dx +(1/2) ∫ (dx/(x^2 −x(√2)+1)) but =(1/(4(√2))) ln∣((x^2 +x(√2)+1)/(x^2 −x(√2) +1))∣ +(1/2) ∫ (dx/((x+((√2)/2))^2 +(1/2))) +(1/2) ∫ (dx/((x−((√2)/2))^2 +(1/2))) but ch. x +((√2)/2) =(1/( (√2))) u give ∫ (dx/((x−((√2)/2))^2 +(1/2))) = 2∫ (1/(1+u^2 )) (du/( (√2))) =(√2) arctan( x(√2)+1) also ch. x−((√2)/2) =(1/( (√2))) u ∫ (dx/((x−((√2)/2))^2 +(1/2))) =2 ∫ (1/(1+u^2 )) (du/( (√2))) =(√2)arctan((√2)x−1) ⇒ ∫ F(x)dx = (1/(4(√2)))ln∣((x^2 +x(√2)+1)/(x^2 −x(√2)+1))∣+((√2)/2) arctan(x(√2)+1)+((√2)/2) arctan(x(√2)−1) +c

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{2}}{x}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\:+\sqrt{\mathrm{2}}{x}\right)} \\ $$$$=\:\frac{{ax}+{b}}{\left({x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}\right)}\:+\frac{{cx}\:+{d}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:\:\:{we}\:{have}\:{F}\left(−{x}\right)={F}\left({x}\right)\:\Rightarrow \\ $$$$\frac{−{ax}+{b}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:+\frac{−{cx}\:+{d}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:={F}\left({x}\right)\:\Rightarrow{c}=−{a}\:{and}\:\:{b}={d}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:−{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:+\frac{−{ax}\:+{b}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\mathrm{2}{b}\:\Rightarrow{b}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{{a}+{b}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:+\frac{−{a}\:+{b}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left({a}+{b}\right)+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(−{a}+{b}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{2}}−\mathrm{2}+\sqrt{\mathrm{2}}\right){a}\:+\left(\mathrm{2}+\sqrt{\mathrm{2}}+\mathrm{2}−\sqrt{\mathrm{2}}\right){b}\:=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{a}\:+\mathrm{4}{b}\:=\mathrm{1}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}{a}\:+\mathrm{2}=\mathrm{1}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:+\:\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:−{x}\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\:\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:+\frac{−{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\int\left\{\:\:\frac{\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\:−\frac{\mathrm{2}{x}−\mathrm{2}\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}+\mathrm{1}}\right\}{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\int\:\:\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}{dx}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\int\:\:\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}{dx} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}+\mathrm{1}}\:{but} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:{ln}\mid\frac{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{2}}\:+\mathrm{1}}\mid\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{\left({x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dx}}{\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:\:{but}\:{ch}. \\ $$$${x}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{u}\:\:{give}\:\int\:\:\:\:\:\frac{{dx}}{\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\mathrm{2}\int\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{2}}\:{arctan}\left(\:{x}\sqrt{\mathrm{2}}+\mathrm{1}\right)\:\:{also}\:{ch}.\:{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{u} \\ $$$$\int\:\:\:\frac{{dx}}{\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:=\mathrm{2}\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\mid\frac{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{2}}+\mathrm{1}}{{x}^{\mathrm{2}} \:−{x}\sqrt{\mathrm{2}}+\mathrm{1}}\mid+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left({x}\sqrt{\mathrm{2}}+\mathrm{1}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left({x}\sqrt{\mathrm{2}}−\mathrm{1}\right)\:+{c}\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Oct/18

∫((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)∫((1+(1/x^2 ))/((x−(1/x))^2 +2))dx−(1/2)∫((1−(1/x^2 ))/((x+(1/x))^2 −2))dx (1/2)∫((d(x−(1/x)))/((x−(1/x))^2 +2))dx−(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

$$\int\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}\:}\right)−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}\:}\right)+{c} \\ $$

Commented by arvinddayama01@gmail.com last updated on 04/Oct/18

$$\mathrm{mind}\:\mathrm{blowing}\:\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Oct/18

$${happiness}\:{latent}\:{in}\:{words}…{thank}\:{you}\:… \\ $$

Commented by arvinddayama01@gmail.com last updated on 04/Oct/18

$$\because\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\mathrm{hahaha}…. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Oct/18