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Question Number 49829 by mhozhez last updated on 11/Dec/18
∫(x^2 /(x^4 +1))dx
$$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{1}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 11/Dec/18
let I = ∫ (x^2 /(x^4 +1))dx vhangement x =(1/t) give I = ∫ (1/(t^2 ( (1/t^4 ) +1))) −(dt/t^2 ) = −∫ (dt/(t^4 (1+(1/t^4 )))) = ∫ (dt/(t^4 +1)) let decompose F(t) =(1/(1+t^4 )) ⇒F(t) = (1/((t^2 +1)^2 −2t^2 )) =(1/((t^2 +(√2)t +1)(t^2 −(√2)t +1))) =((at +b)/(t^2 +(√2)t +1)) + ((ct +d)/(t^2 −(√2) t +1)) we have F(−t)=F(t) ⇒ ((−at +b)/(t^2 −(√2)t +1)) +((−ct +d)/(t^2 +(√2)t +1)) =F(t) ⇒c =−a and b=d ⇒ F(t) = ((at +b)/(t^2 +(√2)t +1)) +((−at +b)/(t^2 −(√2)t +1)) F(0) =1 = 2b ⇒b=(1/2) F(1) =(1/2)= ((a+b)/(2+(√2))) +((−a+b)/(2−(√2))) ⇒(((2−(√2))(a+b)+(2+(√2))(−a+b))/2) =(1/2) ⇒ (2−(√2)−2−(√2))a +(2−(√2) +2+(√2))b =1 ⇒ −2(√2)a +4 .(1/2) =1 ⇒−2(√2)a =−1 ⇒a =(1/(2(√2))) ⇒ F(x) =(((1/(2(√2)))t +(1/2))/(t^2 +(√2)t +1)) +((−(1/(2(√2)))t +(1/2))/(t^2 −(√2)t +1)) ⇒ ∫ F(x)dx = (1/(2(√2))) ∫ ((t +(√2))/(t^2 +(√2)t +1))dt −(1/(2(√2))) ∫ ((t −(√2))/(t^2 −(√2)t +1)) dt =H −K 2(√2)H = (1/2) ∫ ((2t +(√2) +(√2))/(t^2 +(√2)t +1)) dt =(1/2)ln(t^2 +(√2)t +1) +((√2)/2) ∫ (dt/(t^2 +(√2)t +1)) ∫ (dt/(t^2 +(√2)t +1)) = ∫ (dt/(t^2 +2 ((√2)/2)t +(1/2) +1−(1/2))) =∫ (dt/((t +((√2)/2))^2 +(1/2))) =_(t+((√2)/2)=(1/( (√2))) u) 2∫ (1/(1+u^2 )) (du/( (√2))) =(√2)arctan(((t(√2)+1)/( (√2)))) +c ⇒ 2(√2)H = (1/2)ln(t^2 +(√2)t +1) +arctan(((t(√2) +1)/( (√2))))+c but t =(1/x) ⇒ H =(1/(4(√2)))ln((1/x^2 ) +((√2)/x) +1) +(1/(2(√2))) arctan((1/x) +(1/( (√2))) ) +c we follow the same manner for calculus of K .
$${let}\:{I}\:=\:\int\:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:{vhangement}\:\:{x}\:=\frac{\mathrm{1}}{{t}}\:\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left(\:\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\:+\mathrm{1}\right)}\:−\frac{{dt}}{{t}^{\mathrm{2}} }\:=\:−\int\:\:\:\frac{{dt}}{{t}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right)}\:=\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }\:\Rightarrow{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)} \\ $$$$=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\:\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:{we}\:{have}\:{F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow \\ $$$$\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:={F}\left({t}\right)\:\Rightarrow{c}\:=−{a}\:{and}\:{b}={d}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\:\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\:\mathrm{2}{b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}=\:\frac{{a}+{b}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:\:+\frac{−{a}+{b}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:\:\Rightarrow\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left({a}+{b}\right)+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left(−{a}+{b}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{2}}−\mathrm{2}−\sqrt{\mathrm{2}}\right){a}\:+\left(\mathrm{2}−\sqrt{\mathrm{2}}\:+\mathrm{2}+\sqrt{\mathrm{2}}\right){b}\:=\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{2}\sqrt{\mathrm{2}}{a}\:\:+\mathrm{4}\:.\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{1}\:\Rightarrow−\mathrm{2}\sqrt{\mathrm{2}}{a}\:=−\mathrm{1}\:\Rightarrow{a}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{t}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{t}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\:\:\frac{{t}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}{dt}\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\frac{{t}\:−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:{dt}\:\:={H}\:−{K} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{H}\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{2}{t}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$$\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:=\:\int\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{t}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:=\int\:\:\frac{{dt}}{\left({t}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=_{{t}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{u}} \:\:\mathrm{2}\int\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}{arctan}\left(\frac{{t}\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:+{c}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{H}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\:+{arctan}\left(\frac{{t}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{c}\:\:\:{but}\:{t}\:=\frac{\mathrm{1}}{{x}}\:\:\Rightarrow \\ $$$${H}\:=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\sqrt{\mathrm{2}}}{{x}}\:+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\right)\:+{c}\:{we}\:{follow}\:{the}\:{same} \\ $$$${manner}\:{for}\:{calculus}\:{of}\:{K}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
∫(dx/(x^2 +(1/x^2 ))) =(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(√2) ))ln{(((x+(1/x))−(√2))/((x+(1/x))+(√2)))}+(1/2)×(1/( (√2)))tan^(−1) (((x−(1/x))/( (√2))))+c pls check...
$$\int\frac{{dx}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\left\{\frac{\left({x}+\frac{\mathrm{1}}{{x}}\right)−\sqrt{\mathrm{2}}}{\left({x}+\frac{\mathrm{1}}{{x}}\right)+\sqrt{\mathrm{2}}}\right\}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$${pls}\:{check}… \\ $$
Commented by mhozhez last updated on 11/Dec/18
thanks sir
$${thanks}\:{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
most welcome...
$${most}\:{welcome}… \\ $$

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