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-gt-0-x-3-px-q-0-has-got-1-real-and-2-conjugated-complex-solutions-u-

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Question Number 68141 by MJS last updated on 06/Sep/19

$$\mathrm{the}2\mathrm{formulas}\mathrm{for}\mathrm{solving}\int \frac{dx}{x^3+px+q}\mathrm{with}$$$$“{\mathrm{nasty}}^{″}\mathrm{solutions}\mathrm{of}{x}^3+px+q=0\mathrm{with}p,q\in \mathbb{R}$$$$$$$$\mathrm{case}1$$$$D=\frac{p^3}{27}+\frac{q^2}{4}>0\Rightarrow x^3+px+q=0\mathrm{has}\mathrm{got}1\mathrm{real}$$$$\mathrm{and}2\mathrm{conjugated}\mathrm{complex}\mathrm{solutions}$$$$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}\wedge v=\sqrt[3]{-\frac{q}{2}-p\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}$$$$x_1=u+v$$$$x_2=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})v$$$$x_3=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})v$$$$\alpha =u+v\wedge \beta =\frac{\sqrt{3}}{2}(u-v)\iff u=\frac{\alpha }{2}+\frac{\beta }{\sqrt{3}}\wedge v=\frac{\alpha }{2}-\frac{\beta }{\sqrt{3}}$$$$x_1=\alpha$$$$x_2=-\frac{\alpha }{2}+\beta \mathrm{i}$$$$x_3=-\frac{\alpha }{2}-\beta \mathrm{i}$$$$\int \frac{dx}{x^3+px+q}=\int \frac{dx}{(x-\alpha )(x^2+\alpha x+\frac{{\alpha }^2+4{\beta }^2}{4})}=$$$$=\frac{1}{9{\alpha }^2+4{\beta }^2}(\int \frac{dx}{(x-\alpha )}-\int \frac{x+2\alpha }{x^2+\alpha x+\frac{{\alpha }^2+4{\beta }^2}{4}}dx)=$$$$=\frac{1}{9{\alpha }^2+4{\beta }^2}(\ln \mid x-\alpha \mid -\frac{1}{2}\ln ({(2x+\alpha )}^2+4{\beta }^2)-\frac{3\alpha }{2\beta }\arctan \frac{2x+\alpha }{2\beta })+C$$$$\dots \mathrm{now}\mathrm{calculate}\mathrm{the}\mathrm{constants}$$$$$$$$\mathrm{case}2$$$$D=\frac{p^3}{27}+\frac{q^2}{4}<0\Rightarrow x^3+px+q=0\mathrm{has}\mathrm{got}3\mathrm{real}\mathrm{solutions}$$$$x_k=\frac{2}{3}\sqrt{-3p}\sin (\frac{2\pi }{3}k+\frac{1}{3}\arcsin \frac{3\sqrt{3}q}{2\sqrt{-p^3}})\mathrm{with}k=0,1,2$$$$\mathrm{let}{x}_1=\alpha ,x_2=\beta ,x_3=\gamma$$$$\int \frac{dx}{x^3+px+q}=\int \frac{dx}{(x-\alpha )(x-\beta )(x-\gamma )}=$$$$=\frac{1}{(\alpha -\beta )(\alpha -\gamma )}\int \frac{dx}{x-\alpha }+\frac{1}{(\beta -\alpha )(\beta -\gamma )}\int \frac{dx}{x-\beta }+\frac{1}{(\gamma -\alpha )(\gamma -\beta )}\int \frac{dx}{x-\gamma }=$$$$=\frac{\ln \mid x-\alpha \mid }{(\alpha -\beta )(\alpha -\gamma )}+\frac{\ln \mid x-\beta \mid }{(\beta -\alpha )(\beta -\gamma )}+\frac{\ln \mid x-\gamma \mid }{(\gamma -\alpha )(\gamma -\beta )}+C$$$$\dots \mathrm{now}\mathrm{calculate}\mathrm{the}\mathrm{constants}$$

Commented by mind is power last updated on 06/Sep/19

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