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Question Number 132799 by Ñï= last updated on 16/Feb/21
∫(dx/(1+x^4 )) =(1/2)∫(dx/(1+ix^2 ))+(1/2)∫(dx/(1−ix^2 )) =(1/2)e^(−i(π/4)) ∫((d(e^(i(π/4)) x))/(1+(e^(i(π/4)) x)^2 ))+(1/2)e^(i(π/4)) ∫((d(e^(−i(π/4)) )x)/(1+(e^(−i(π/4)) x)^2 )) =(1/2)e^(−i(π/4)) tan^(−1) (e^(i(π/4)) x)+(1/2)e^(i(π/4)) tan^(−1) (e^(−i(π/4)) x)+C =Question is how to continue finish it?
$$\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}+{ix}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}−{ix}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \int\frac{{d}\left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)}{\mathrm{1}+\left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \int\frac{{d}\left({e}^{−{i}\frac{\pi}{\mathrm{4}}} \right){x}}{\mathrm{1}+\left({e}^{−{i}\frac{\pi}{\mathrm{4}}} {x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} {tan}^{−\mathrm{1}} \left({e}^{{i}\frac{\pi}{\mathrm{4}}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} {tan}^{−\mathrm{1}} \left({e}^{−{i}\frac{\pi}{\mathrm{4}}} {x}\right)+{C} \\ $$$$={Question}\:{is}\:{how}\:{to}\:{continue}\:{finish}\:{it}? \\ $$
Commented by Ar Brandon last updated on 16/Feb/21
x^4 +1=(x^2 +1)^2 −2x^2 =(x^2 +1)^2 −((√2)x)^2 =(x^2 −(√2)x+1)(x^2 +(√2)x+1) (1/(x^4 +1))=((ax+b)/(x^2 −(√2)x+1))+((cx+d)/(x^2 +(√2)x+1)) a+c=0, (√2)a+b−(√2)c+d=0, a+(√2)b+c−(√2)d=0, b+d=1 b=(1/2)=d, a=−(1/(2(√2))), c=(1/(2(√2)))
$$\mathrm{x}^{\mathrm{4}} +\mathrm{1}=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}+\frac{\mathrm{cx}+\mathrm{d}}{\mathrm{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}} \\ $$$$\mathrm{a}+\mathrm{c}=\mathrm{0},\:\sqrt{\mathrm{2}}\mathrm{a}+\mathrm{b}−\sqrt{\mathrm{2}}\mathrm{c}+\mathrm{d}=\mathrm{0},\:\mathrm{a}+\sqrt{\mathrm{2}}\mathrm{b}+\mathrm{c}−\sqrt{\mathrm{2}}\mathrm{d}=\mathrm{0},\:\mathrm{b}+\mathrm{d}=\mathrm{1} \\ $$$$\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{d},\:\mathrm{a}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}},\:\mathrm{c}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by MJS_new last updated on 16/Feb/21
arctan (a+bi) = =(1/2)(πsign a +arctan ((b−1)/a) −arctan ((b+1)/a))+(i/4)ln ((a^2 +(b+1)^2 )/(a^2 +(b−1)^2 )) that′s why I′d rather not...
$$\mathrm{arctan}\:\left({a}+{b}\mathrm{i}\right)\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi\mathrm{sign}\:{a}\:+\mathrm{arctan}\:\frac{{b}−\mathrm{1}}{{a}}\:−\mathrm{arctan}\:\frac{{b}+\mathrm{1}}{{a}}\right)+\frac{\mathrm{i}}{\mathrm{4}}\mathrm{ln}\:\frac{{a}^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{why}\:\mathrm{I}'\mathrm{d}\:\mathrm{rather}\:\mathrm{not}… \\ $$
Answered by MJS_new last updated on 16/Feb/21
∫(dx/(x^4 +1))=(1/4)∫(((√2)x+2)/(x^2 +(√2)x+1))−(((√2)x−2)/(x^2 −(√2)x+1))dx= =∫((2x+(√2))/(4(√2)(x^2 +(√2)x+1)))+(1/(4(x^2 +(√2)x+1)))−((2x−(√2))/(4(√2)(x^2 +(√2)x+1)))+(1/(4(x^2 −(√2)x+1)))dx= =((√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) +C
$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\sqrt{\mathrm{2}}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}−\frac{\sqrt{\mathrm{2}}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}= \\ $$$$=\int\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)}−\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+{C} \\ $$
Commented by Ñï= last updated on 16/Feb/21
I want to know how to use arctan fuction to get it back. This method I already get it.sir.
$${I}\:{want}\:{to}\:{know}\:{how}\:{to}\:{use}\:\mathrm{arc}{tan}\:{fuction}\:{to}\:{get}\:{it}\:{back}. \\ $$$${This}\:{method}\:{I}\:{already}\:{get}\:{it}.{sir}. \\ $$
Answered by Ar Brandon last updated on 16/Feb/21
∫(dx/(x^4 +1))=(1/2)∫(((x^2 +1)−(x^2 −1))/(x^4 +1))dx=(1/2)∫((x^2 +1)/(x^4 +1))dx−(1/2)∫((x^2 −1)/(x^4 +1))dx =(1/2)∫((1+(1/x^2 ))/(x^2 +(1/x^2 )))dx−(1/2)∫((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)∫(du/(u^2 +2))−(1/2)∫(dv/(v^2 −2))=((tan^(−1) (u/(√2)))/(2(√2)))+((tanh^(−1) (v/(√2)))/(2(√2)))+C =(1/(2(√2)))tan^(−1) (((x^2 −1)/( (√2)x)))+(1/(4(√2)))ln∣((x^2 +x+1)/(x^2 −x+1))∣+C
$$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{2}}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{2}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{2}}=\frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{u}/\sqrt{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{v}/\sqrt{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}+\mathcal{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}}\right)+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\mid+\mathrm{C} \\ $$
Answered by mathmax by abdo last updated on 16/Feb/21
for that find arctan(x+iy)...
$$\mathrm{for}\:\mathrm{that}\:\mathrm{find}\:\mathrm{arctan}\left(\mathrm{x}+\mathrm{iy}\right)… \\ $$

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