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


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 68141 by MJS last updated on 06/Sep/19

$the 2 formulas for solving \int \frac{d x}{x^{3} + p x + q} with$ $‘ ‘ {nasty}^{″} solutions of x^{3} + p x + q = 0 with p, q \in R$ $case 1$ $D = \frac{p^{3}}{27} + \frac{q^{2}}{4} > 0 \Rightarrow x^{3} + p x + q = 0 has got 1 real$ $and 2 conjugated complex solutions$ $u = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} \land v = \sqrt[3]{- \frac{q}{2} - p \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}}$ $x_{1} = u + v$ $x_{2} = (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v$ $x_{3} = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v$ $α = u + v \land β = \frac{\sqrt{3}}{2} (u - v) \Leftrightarrow u = \frac{α}{2} + \frac{β}{\sqrt{3}} \land v = \frac{α}{2} - \frac{β}{\sqrt{3}}$ $x_{1} = α$ $x_{2} = - \frac{α}{2} + β i$ $x_{3} = - \frac{α}{2} - β i$ $\int \frac{d x}{x^{3} + p x + q} = \int \frac{d x}{(x - α) (x^{2} + α x + \frac{α^{2} + 4 β^{2}}{4})} =$ $= \frac{1}{9 α^{2} + 4 β^{2}} (\int \frac{d x}{(x - α)} - \int \frac{x + 2 α}{x^{2} + α x + \frac{α^{2} + 4 β^{2}}{4}} d x) =$ $= \frac{1}{9 α^{2} + 4 β^{2}} (\ln ∣ x - α ∣ - \frac{1}{2} \ln ({(2 x + α)}^{2} + 4 β^{2}) - \frac{3 α}{2 β} \arctan \frac{2 x + α}{2 β}) + C$ $. . . now calculate the constants$ $case 2$ $D = \frac{p^{3}}{27} + \frac{q^{2}}{4} < 0 \Rightarrow x^{3} + p x + q = 0 has got 3 real solutions$ $x_{k} = \frac{2}{3} \sqrt{- 3 p} \sin (\frac{2 π}{3} k + \frac{1}{3} \arcsin \frac{3 \sqrt{3} q}{2 \sqrt{- p^{3}}}) with k = 0, 1, 2$ $let x_{1} = α, x_{2} = β, x_{3} = γ$ $\int \frac{d x}{x^{3} + p x + q} = \int \frac{d x}{(x - α) (x - β) (x - γ)} =$ $= \frac{1}{(α - β) (α - γ)} \int \frac{d x}{x - α} + \frac{1}{(β - α) (β - γ)} \int \frac{d x}{x - β} + \frac{1}{(γ - α) (γ - β)} \int \frac{d x}{x - γ} =$ $= \frac{\ln ∣ x - α ∣}{(α - β) (α - γ)} + \frac{\ln ∣ x - β ∣}{(β - α) (β - γ)} + \frac{\ln ∣ x - γ ∣}{(γ - α) (γ - β)} + C$ $. . . now calculate the constants$
Commented by mind is power last updated on 06/Sep/19

$t h a n k y o u f o r t h i s w o r c k!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum