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Question Number 68141 by MJS last updated on 06/Sep/19
the 2 formulas for solving ∫(dx/(x^3 +px+q)) with “nasty” solutions of x^3 +px+q=0 with p, q ∈R case 1 D=(p^3 /(27))+(q^2 /4)>0 ⇒ x^3 +px+q=0 has got 1 real and 2 conjugated complex solutions u=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) ∧v=((−(q/2)−p(√((p^3 /(27))+(q^2 /4)))))^(1/3) x_1 =u+v x_2 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v x_3 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v α=u+v∧β=((√3)/2)(u−v) ⇔ u=(α/2)+(β/( (√3)))∧v=(α/2)−(β/( (√3))) x_1 =α x_2 =−(α/2)+βi x_3 =−(α/2)−βi ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x^2 +αx+((α^2 +4β^2 )/4))))= =(1/(9α^2 +4β^2 ))(∫(dx/((x−α)))−∫((x+2α)/(x^2 +αx+((α^2 +4β^2 )/4)))dx)= =(1/(9α^2 +4β^2 ))(ln ∣x−α∣ −(1/2)ln ((2x+α)^2 +4β^2 ) −((3α)/(2β))arctan ((2x+α)/(2β))) +C ...now calculate the constants case 2 D=(p^3 /(27))+(q^2 /4)<0 ⇒ x^3 +px+q=0 has got 3 real solutions x_k =(2/3)(√(−3p)) sin (((2π)/3)k+(1/3)arcsin ((3(√3)q)/(2(√(−p^3 ))))) with k=0, 1, 2 let x_1 =α, x_2 =β, x_3 =γ ∫(dx/(x^3 +px+q))=∫(dx/((x−α)(x−β)(x−γ)))= =(1/((α−β)(α−γ)))∫(dx/(x−α))+(1/((β−α)(β−γ)))∫(dx/(x−β))+(1/((γ−α)(γ−β)))∫(dx/(x−γ))= =((ln ∣x−α∣)/((α−β)(α−γ)))+((ln ∣x−β∣)/((β−α)(β−γ)))+((ln ∣x−γ∣)/((γ−α)(γ−β)))+C ...now calculate the constants
$$\mathrm{the}\:\mathrm{2}\:\mathrm{formulas}\:\mathrm{for}\:\mathrm{solving}\:\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}\:\mathrm{with} \\ $$$$“\mathrm{nasty}''\:\mathrm{solutions}\:\mathrm{of}\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{with}\:{p},\:{q}\:\in\mathbb{R} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{1}\:\mathrm{real} \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}\wedge{v}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−{p}\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$${x}_{\mathrm{1}} ={u}+{v} \\ $$$${x}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$${x}_{\mathrm{3}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\alpha={u}+{v}\wedge\beta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({u}−{v}\right)\:\Leftrightarrow\:{u}=\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\:\sqrt{\mathrm{3}}}\wedge{v}=\frac{\alpha}{\mathrm{2}}−\frac{\beta}{\:\sqrt{\mathrm{3}}} \\ $$$${x}_{\mathrm{1}} =\alpha \\ $$$${x}_{\mathrm{2}} =−\frac{\alpha}{\mathrm{2}}+\beta\mathrm{i} \\ $$$${x}_{\mathrm{3}} =−\frac{\alpha}{\mathrm{2}}−\beta\mathrm{i} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}=\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}^{\mathrm{2}} +\alpha{x}+\frac{\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }{\mathrm{4}}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }\left(\int\frac{{dx}}{\left({x}−\alpha\right)}−\int\frac{{x}+\mathrm{2}\alpha}{{x}^{\mathrm{2}} +\alpha{x}+\frac{\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }{\mathrm{4}}}{dx}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} }\left(\mathrm{ln}\:\mid{x}−\alpha\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\left(\mathrm{2}{x}+\alpha\right)^{\mathrm{2}} +\mathrm{4}\beta^{\mathrm{2}} \right)\:−\frac{\mathrm{3}\alpha}{\mathrm{2}\beta}\mathrm{arctan}\:\frac{\mathrm{2}{x}+\alpha}{\mathrm{2}\beta}\right)\:+{C} \\ $$$$…\mathrm{now}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{constants} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{has}\:\mathrm{got}\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions} \\ $$$${x}_{{k}} =\frac{\mathrm{2}}{\mathrm{3}}\sqrt{−\mathrm{3}{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}{k}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}{q}}{\mathrm{2}\sqrt{−{p}^{\mathrm{3}} }}\right)\:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\mathrm{let}\:{x}_{\mathrm{1}} =\alpha,\:{x}_{\mathrm{2}} =\beta,\:{x}_{\mathrm{3}} =\gamma \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{3}} +{px}+{q}}=\int\frac{{dx}}{\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)}= \\ $$$$=\frac{\mathrm{1}}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)}\int\frac{{dx}}{{x}−\alpha}+\frac{\mathrm{1}}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)}\int\frac{{dx}}{{x}−\beta}+\frac{\mathrm{1}}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)}\int\frac{{dx}}{{x}−\gamma}= \\ $$$$=\frac{\mathrm{ln}\:\mid{x}−\alpha\mid}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)}+\frac{\mathrm{ln}\:\mid{x}−\beta\mid}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)}+\frac{\mathrm{ln}\:\mid{x}−\gamma\mid}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)}+{C} \\ $$$$…\mathrm{now}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{constants} \\ $$
Commented by mind is power last updated on 06/Sep/19
thank you for this worck!
$${thank}\:{you}\:{for}\:{this}\:{worck}! \\ $$

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