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Question Number 125114 by bramlexs22 last updated on 08/Dec/20

$$solve\int \frac{dx}{{(x^3-1)}^2}?$$

Commented by liberty last updated on 08/Dec/20

$${Ostrogradsky}^{\prime }smethod\left(ProfsirMJS\right)$$

Answered by liberty last updated on 08/Dec/20

$$equationhastheform\int \frac{dx}{{(x^3-1)}^2}=\frac{{ax}^2+bx+c}{x^3-1}+\int \frac{{ex}^2+fx+g}{x^3-1}dx$$$$differentiatingbothsidewehave$$$$\frac{1}{{(x^3-1)}^2}=\frac{(x^3-1)(2ax+b)-({ax}^2+bx+c)3x^2}{{(x^3-1)}^2}+\frac{{ex}^2+fx+g}{x^3-1}$$$$$$$$equatingthecoefficientsgives$$$$e=0,a=0,c=0,b=-\frac{1}{3},f=0,g=-\frac{2}{3}$$$$I=\int \frac{dx}{{(x^3-1)}^2}=\frac{-\frac{1}{3}x}{x^3-1}+\int \frac{\frac{-2}{3}}{x^3-1}dx$$$$=\frac{-x}{3(x^3-1)}+\int [\frac{-\frac{2}{9}}{x-1}+\frac{\frac{2}{9}x+\frac{4}{9}}{x^2+x+1}]dx$$$$=-\frac{x}{3(x^3-1)}-\frac{2}{9}\ln \mid x-1\mid +\frac{1}{9}\ln \mid x^2+x+1\mid$$$$+\frac{2\sqrt{3}}{9}\mathrm{arc}\tan \left(\frac{2x+1}{\sqrt{3}}\right)+c$$

Commented by bramlexs22 last updated on 08/Dec/20

$$thankyousir$$

Commented by MJS_new last updated on 08/Dec/20

$$\mathrm{yesss}!$$

Commented by bramlexs22 last updated on 08/Dec/20

does anyone have a pdf file about this method

Commented by MJS_new last updated on 08/Dec/20

$$\mathrm{search}\mathrm{for}‘‘\mathrm{Ostrogradski}{\mathrm{Method}}^{″}\mathrm{on}\mathrm{the}$$$$\mathrm{www},\mathrm{plenty}\mathrm{of}\mathrm{explanations}...$$

Commented by Ar Brandon last updated on 08/Dec/20

well detailed��

Answered by mathmax by abdo last updated on 08/Dec/20

$$\mathrm{A}=\int \frac{\mathrm{dx}}{{({\mathrm{x}}^3-1)}^2}\mathrm{let}\mathrm{f}\left(\lambda \right)=\int \frac{\mathrm{dx}}{{\mathrm{x}}^3-{\lambda }^3}(\lambda >0)\mathrm{we}\mathrm{have}$$$${\mathrm{f}}^{{}^{\prime }}\left(\lambda \right)=-\int \frac{3{\lambda }^2}{{({\mathrm{x}}^3-{\lambda }^3)}^2}\mathrm{dx}\Rightarrow \int \frac{\mathrm{dx}}{{\mathrm{x}}^3-{\lambda }^3}=-\frac{1}{3{\lambda }^2}{\mathrm{f}}^{{}^{\prime }}\left(\lambda \right)\mathrm{and}$$$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^3-1}=-\frac{1}{3}{\mathrm{f}}^{{}^{\prime }}\left(1\right)\mathrm{let}\mathrm{explicit}\mathrm{f}\left(\lambda \right)\mathrm{let}\mathrm{x}=\lambda \mathrm{u}\Rightarrow$$$$\mathrm{f}\left(\lambda \right)=\int \frac{\lambda \mathrm{du}}{{\lambda }^3({\mathrm{u}}^3-1)}=\frac{1}{{\lambda }^2}\int \frac{\mathrm{du}}{{\mathrm{u}}^3-1}\mathrm{and}$$$$\mathrm{g}\left(\mathrm{u}\right)=\frac{1}{{\mathrm{u}}^3-1}=\frac{1}{(\mathrm{u}-1)({\mathrm{u}}^2+\mathrm{u}+1)}=\frac{\mathrm{a}}{\mathrm{u}-1}+\frac{\mathrm{bu}+\mathrm{c}}{{\mathrm{u}}^2+\mathrm{u}+1}$$$$\mathrm{a}=\frac{1}{3},{\lim }_{\mathrm{u}\to +\mathrm{\infty }}\mathrm{uG}\left(\mathrm{u}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-\frac{1}{3}$$$$\mathrm{g}\left(\mathrm{o}\right)=-1=-\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=-1+\mathrm{a}=-1+\frac{1}{3}=-\frac{2}{3}\Rightarrow$$$$\mathrm{g}\left(\mathrm{u}\right)=\frac{1}{3(\mathrm{u}-1)}+\frac{-\frac{1}{3}\mathrm{u}-\frac{2}{3}}{{\mathrm{u}}^2+\mathrm{u}+1}=\frac{1}{3(\mathrm{u}-1)}-\frac{1}{3}\times \frac{\mathrm{u}+2}{{\mathrm{u}}^2+\mathrm{u}+1}$$$$=\frac{1}{3(\mathrm{u}-1)}-\frac{1}{6}\times \frac{2\mathrm{u}+1+3}{{\mathrm{u}}^2+\mathrm{u}+1}\Rightarrow \int \mathrm{g}\left(\mathrm{u}\right)\mathrm{du}=\frac{1}{3}\ln \mid \mathrm{u}-1\mid$$$$-\frac{1}{6}\ln ({\mathrm{u}}^2+\mathrm{u}+1)-\frac{1}{2}\int \frac{\mathrm{du}}{{(\mathrm{u}+\frac{1}{2})}^2+\frac{3}{4}}(\to \mathrm{u}+\frac{1}{2}=\frac{\sqrt{3}}{2}\mathrm{z})\Rightarrow$$$$\int \frac{\mathrm{du}}{{(\mathrm{u}+\frac{1}{2})}^2+\frac{3}{4}}=\frac{\sqrt{3}}{2}\times \frac{4}{3}\int \frac{\mathrm{dz}}{{\mathrm{z}}^2+1}=\frac{2}{\sqrt{3}}\arctan \left(\frac{2\mathrm{u}+1}{\sqrt{3}}\right)\Rightarrow$$$$\int \mathrm{g}\left(\mathrm{u}\right)\mathrm{du}=\frac{1}{3}\ln \mid \mathrm{u}-1\mid -\frac{1}{6}\ln ({\mathrm{u}}^2+\mathrm{u}+1)-\frac{1}{\sqrt{3}}\arctan \left(\frac{2\mathrm{u}+1}{\sqrt{3}}\right)+\mathrm{c}$$$$\mathrm{f}\left(\lambda \right)=\frac{1}{{\lambda }^2}\{\frac{1}{3}\ln \mid \frac{\mathrm{x}}{\lambda }-1\mid -\frac{1}{6}\ln (\frac{{\mathrm{x}}^2}{{\lambda }^2}+\frac{\mathrm{x}}{\lambda }+1)-\frac{1}{\sqrt{3}}\arctan \left(\frac{\frac{2\mathrm{x}}{\lambda }+1}{\sqrt{3}}\right)+\mathrm{c}\Rightarrow$$$$\mathrm{now}\mathrm{its}\mathrm{eazy}\mathrm{to}\mathrm{calculate}{\mathrm{f}}^{{}^{\prime }}\left(\lambda \right)\mathrm{and}{\mathrm{f}}^{{}^{\prime }}\left(1\right)...\mathrm{be}\mathrm{continued}...$$$$$$

Commented by mathmax by abdo last updated on 08/Dec/20

$$\int \frac{\mathrm{dx}}{{({\mathrm{x}}^3-{\lambda }^3)}^2}=-\frac{1}{3{\lambda }^2}{\mathrm{f}}^{{}^{\prime }}\left(\lambda \right)\Rightarrow \int \frac{\mathrm{dx}}{{({\mathrm{x}}^3-1)}^2}=-\frac{1}{3}{\mathrm{f}}^{{}^{\prime }}\left(1\right)$$$$\left(\mathrm{eror}\mathrm{of}\mathrm{typo}\right)$$

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