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Question Number 156170 by MathSh last updated on 08/Oct/21

Answered by mindispower last updated on 09/Oct/21

Hello sir withe resodue[theorm is diractly ((p(x))/(q(x)))= deg q>deg(p)+1 in this case

$${Hello}\:{sir}\:{withe}\:{resodue}\left[{theorm}\:{is}\:{diractly}\:\right. \\ $$$$\frac{{p}\left({x}\right)}{{q}\left({x}\right)}=\:{deg}\:{q}>{deg}\left({p}\right)+\mathrm{1}\:{in}\:{this}\:{case} \\ $$$$ \\ $$$$ \\ $$

Commented by MathSh last updated on 09/Oct/21

Hola dear Ser, thank you how, if possible please

$$\mathrm{Hola}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{how},\:\mathrm{if}\:\mathrm{possible}\:\mathrm{please} \\ $$

Answered by MJS_new last updated on 09/Oct/21

∫(dx/((x^2 +x+1)(a^2 x^2 +1)))= =(1/(a^4 −a^2 +1))∫((a^2 x+1)/(x^2 +x+1))dx−(a^2 /(a^4 −a^2 +1))∫((a^2 x−a^2 +1)/(a^2 x^2 +1))dx (1/(a^4 −a^2 +1))∫((a^2 x+1)/(x^2 +x+1))dx= =(a^2 /(2(a^4 −a^2 +1)))∫((2x+1)/(x^2 +x+1))dx−((a^2 −2)/(2(a^4 −a^2 +1))∫(dx/(x^2 +x+1))= =(a^2 /(2(a^4 −a^2 +1)))ln (x^2 +x+1) −(((√3)(a^2 −2))/(3(a^4 −a^2 +1)))arctan (((√3)(2x+1))/3) −(a^2 /(a^4 −a^2 +1))∫((a^2 x−a^2 +1)/(a^2 x^2 +1))dx= =−(a^4 /(a^4 −a^2 +1))∫(x/(a^2 x^2 +1))dx+((a^2 (a^2 −1))/(a^4 −a^2 +1))∫(dx/(a^2 x^2 +1))= =−(a^2 /(2(a^4 −a^2 +1)))ln (a^2 x^2 +1) +((a(a^2 −1))/(a^4 −a^2 +1))arctan (ax) inserting the borders I get Ω(a)= { (((1/(a^4 −a^2 +1))((π/(18))(9a^3 −2(√3)a^2 −9a+4(√3))−a^2 ln a)∧a>0)),((((2(√3))/9)π∧a=0)),((−(1/(a^4 −a^2 +1))((π/(18))(9a^3 +2(√3)a^2 −9a−4(√3))+a^2 ln (−a))∧a<0)) :}

$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\int\frac{{a}^{\mathrm{2}} {x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\int\frac{{a}^{\mathrm{2}} {x}−{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\int\frac{{a}^{\mathrm{2}} {x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}\right)}\int\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{{a}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}\right.}\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}= \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:−\frac{\sqrt{\mathrm{3}}\left({a}^{\mathrm{2}} −\mathrm{2}\right)}{\mathrm{3}\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$−\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\int\frac{{a}^{\mathrm{2}} {x}−{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$=−\frac{{a}^{\mathrm{4}} }{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\int\frac{{x}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} −\mathrm{1}\right)}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\int\frac{{dx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{ln}\:\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)\:+\frac{{a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{arctan}\:\left({ax}\right) \\ $$$$\mathrm{inserting}\:\mathrm{the}\:\mathrm{borders}\:\mathrm{I}\:\mathrm{get} \\ $$$$\Omega\left({a}\right)=\begin{cases}{\frac{\mathrm{1}}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\left(\frac{\pi}{\mathrm{18}}\left(\mathrm{9}{a}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{3}}{a}^{\mathrm{2}} −\mathrm{9}{a}+\mathrm{4}\sqrt{\mathrm{3}}\right)−{a}^{\mathrm{2}} \mathrm{ln}\:{a}\right)\wedge{a}>\mathrm{0}}\\{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\pi\wedge{a}=\mathrm{0}}\\{−\frac{\mathrm{1}}{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} +\mathrm{1}}\left(\frac{\pi}{\mathrm{18}}\left(\mathrm{9}{a}^{\mathrm{3}} +\mathrm{2}\sqrt{\mathrm{3}}{a}^{\mathrm{2}} −\mathrm{9}{a}−\mathrm{4}\sqrt{\mathrm{3}}\right)+{a}^{\mathrm{2}} \mathrm{ln}\:\left(−{a}\right)\right)\wedge{a}<\mathrm{0}}\end{cases} \\ $$

Commented by MathSh last updated on 09/Oct/21

Perfect my dear Ser, thank you

$$\mathrm{Perfect}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

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