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Question Number 130287 by naka3546 last updated on 24/Jan/21
∫ (x/(x^4 +1)) dx = ?
$$\int\:\frac{{x}}{{x}^{\mathrm{4}} +\mathrm{1}}\:{dx}\:\:=\:\:? \\ $$
Answered by mathmax by abdo last updated on 24/Jan/21
I =∫ ((xdx)/(x^4 +1)) let decompose F(x)=(x/(x^4 +1)) ⇒ F(x)=(x/((x^2 +1)^2 −2x^2 ))=(x/((x^2 +1−(√2)x)(x^(2 ) +1+(√2)x))) =((ax+b)/(x^2 −(√2)x +1))+((mx+n)/(x^2 +(√2)x+1)) we see[that F(x)=(1/(2(√2)))((1/(x^2 −(√2)x+1))−(1/(x^2 +(√2)x+1))) ⇒∫ F(x)dx=(1/(2(√2)))∫ (dx/(x^2 −2((√2)/2)x+(1/2)+(1/2)))−(1/(2(√2)))∫ (dx/(x^2 +2((√2)/2)x+(1/2)+(1/2))) =(1/(2(√2)))∫ (dx/((x−(1/( (√2))))^2 +(1/2)))(→x−(1/( (√2)))=(u/( (√2))))−(1/(2(√2)))∫ (dx/((x+(1/( (√2))))^2 +(1/2)))(→x+(1/( (√2)))=(v/( (√2)))) =(1/(2(√2)))∫ (du/( (√2)((1/2)(u^2 +1))))−(1/(2(√2)))∫ (dv/( (√2)(1/2)(v^2 +1))) =(1/2)∫ (du/(u^2 +1))−(1/2)∫ (dv/(v^(2 ) +1)) =(1/2)arctan(u)−(1/2)arctanv +C =(1/2)arctan((√2)x−1)−(1/2)arctan((√2)x+1) +C
$$\mathrm{I}\:=\int\:\:\frac{\mathrm{xdx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} }=\frac{\mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{2}}\mathrm{x}\right)\left(\mathrm{x}^{\mathrm{2}\:} +\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{x}\right)} \\ $$$$=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}\:+\mathrm{1}}+\frac{\mathrm{mx}+\mathrm{n}}{\mathrm{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}\:\mathrm{we}\:\mathrm{see}\left[\mathrm{that}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}\right)\right. \\ $$$$\Rightarrow\int\:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\left(\rightarrow\mathrm{x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{u}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\left(\rightarrow\mathrm{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{v}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\frac{\mathrm{du}}{\:\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)\right)}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\frac{\mathrm{dv}}{\:\sqrt{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{v}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}\:} +\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{u}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctanv}\:+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)\:+\mathrm{C} \\ $$$$ \\ $$
Commented by naka3546 last updated on 24/Jan/21
thanks, sir . How about ∫ (x/(x^4 +3)) dx ?
$$\mathrm{thanks},\:{sir}\:. \\ $$$$ \\ $$$${How}\:\:{about}\:\:\int\:\:\frac{{x}}{{x}^{\mathrm{4}} +\mathrm{3}}\:{dx}\:\:? \\ $$
Commented by Ar Brandon last updated on 24/Jan/21
∫(x/(x^4 +3))dx=(1/2)∫(du/(u^2 +3))=((tan^(−1) (x^2 /(√3)))/(2(√3)))+C
$$\int\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{4}} +\mathrm{3}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{3}}=\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} /\sqrt{\mathrm{3}}\right)}{\mathrm{2}\sqrt{\mathrm{3}}}+\mathcal{C} \\ $$
Commented by mathmax by abdo last updated on 24/Jan/21
∫ ((xdx)/(x^4 +3)) =∫ ((xdx)/(3((x^4 /3)+1))) =∫ ((xdx)/(3(((x/( (^4 (√3)))))^4 +1))) =_((x/((^4 (√3))))=t) (1/3)∫ (((^4 (√3))t)/(1+t^4 ))(^4 (√3))dt =(3^(2/4) /3) ∫ ((tdt)/(1+t^4 )) =....
$$\int\:\:\frac{\mathrm{xdx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{3}}\:=\int\:\:\frac{\mathrm{xdx}}{\mathrm{3}\left(\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{3}}+\mathrm{1}\right)}\:=\int\:\:\frac{\mathrm{xdx}}{\mathrm{3}\left(\left(\frac{\mathrm{x}}{\:\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)}\right)^{\mathrm{4}} +\mathrm{1}\right)} \\ $$$$=_{\frac{\mathrm{x}}{\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)}=\mathrm{t}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\frac{\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)\mathrm{dt}\:=\frac{\mathrm{3}^{\frac{\mathrm{2}}{\mathrm{4}}} }{\mathrm{3}}\:\int\:\:\frac{\mathrm{tdt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:=…. \\ $$
Commented by mathmax by abdo last updated on 26/Jan/21
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$
Answered by Ar Brandon last updated on 24/Jan/21
I=∫(x/(x^4 +1))dx , x^2 =u =(1/2)∫(du/(u^2 +1))=((tan^(−1) (u))/2)+C =((tan^(−1) (x^2 ))/2)+C
$$\mathcal{I}=\int\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\mathrm{dx}\:,\:\mathrm{x}^{\mathrm{2}} =\mathrm{u} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{u}\right)}{\mathrm{2}}+\mathcal{C} \\ $$$$\:\:\:=\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{2}}+\mathcal{C} \\ $$
Answered by mnjuly1970 last updated on 24/Jan/21
x^2 =u xdx=(du/2) φ=∫(du/(2(u^2 +1)))=(1/2) tan^(−1) (x^2 )+C...
$${x}^{\mathrm{2}} ={u} \\ $$$${xdx}=\frac{{du}}{\mathrm{2}} \\ $$$$\phi=\int\frac{{du}}{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{C}… \\ $$

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