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Question Number 210396 by jirodomo last updated on 08/Aug/24

Commented by jirodomo last updated on 08/Aug/24

$$\mathrm{applying}\mathrm{partial}\mathrm{fraction}\mathrm{decomposition}$$$$\mathrm{straight}\mathrm{away}\mathrm{seems}\mathrm{impossible}\mathrm{here}.$$$$\mathrm{is}\mathrm{there}\mathrm{other}\mathrm{way}?\mathrm{im}\mathrm{stuck}\mathrm{for}\mathrm{a}\mathrm{few}$$$$\mathrm{months}\mathrm{now}.$$

Commented by jhlombardi last updated on 08/Aug/24

divide de polinmio of the numbrer by the denominator polinimio

Answered by Frix last updated on 08/Aug/24

$${\mathrm{Where}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{issue}?$$$$x^3-1=(x-1)(x^2+x+1)$$$$\Rightarrow$$$$\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{{(x^3-1)}^2}dx=$$$$=\int (\frac{2(2x+1)}{{(x^2+x+1)}^2}+\frac{2x+1}{x^2+x+1}-\frac{1}{{(x-1)}^2}+\frac{1}{x-1})dx=$$$$...$$$$=\frac{x^2-x+3}{x^3-1}+\ln \mid x^3-1\mid +C$$

Commented by jirodomo last updated on 08/Aug/24

$$\mathrm{How}\mathrm{did}\mathrm{you}\mathrm{find}\mathrm{the}\mathrm{nominators}\mathrm{for}\mathrm{those}\mathrm{fractions}?$$$$\mathrm{my}\mathrm{issue}\mathrm{is}\mathrm{let}\mathrm{the}\mathrm{function}\mathrm{be}$$$$\frac{Ax+B}{{(x^2+x+1)}^2}+\frac{C}{{(x-1)}^2}+\frac{Dx+E}{x^2+x+1}+\frac{F}{x-1}$$$$\mathrm{and}\mathrm{I}\mathrm{ended}\mathrm{up}\mathrm{getting}\mathrm{equation}\mathrm{system}$$$$\mathrm{with}6\mathrm{variables}.\mathrm{Its}\mathrm{mind}\mathrm{numbing}.$$$$\mathrm{or}\mathrm{did}\mathrm{you}\mathrm{also}\mathrm{go}\mathrm{through}\mathrm{this}\mathrm{way}?$$

Commented by Frix last updated on 08/Aug/24

$$\mathrm{Search}\mathrm{the}\mathrm{www}\mathrm{for}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}$$$$\mathrm{to}\mathrm{see}\mathrm{the}\mathrm{proof}.$$$$\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{{(x^3-1)}^2}dx=$$$$=\frac{{Ax}^2+Bx+C}{x^3-1}+\int \frac{{Dx}^2+Ex+F}{x^3-1}dx$$$$\mathrm{Now}\mathrm{differentiate}\mathrm{both}\mathrm{sides}:$$$$\frac{3x^5-x^4+2x^3-12x^2-2x+1}{{(x^3-1)}^2}=$$$$=-\frac{{Ax}^4+2{Bx}^3+3{Cx}^2+2Ax+B}{{(x^3-1)}^2}+$$$$+\frac{{Dx}^2+Ex+F}{x^3-1}$$$$\mathrm{Multiplicate}\mathrm{with}{(x^3-1)}^2\mathrm{and}\mathrm{transform}$$$$(D-3)x^5$$$$-(A-E-1)x^4$$$$-(2B-F+2)x^3$$$$-(3C+D-12)x^2$$$$-(2A+E-2)x$$$$-(B+F+1)$$$$=0$$$$\Rightarrow A=1,B=-1,C=D=3,E=F=0$$$$\mathrm{Now}\mathrm{we}\mathrm{have}$$$$\int \frac{3x^5-x^4+2x^3-12x^2-2x+1}{{(x^3-1)}^2}dx=$$$$=\frac{x^2-x+3}{x^3-1}+3\int \frac{x^2}{x^3-1}dx$$

Commented by jirodomo last updated on 09/Aug/24

$$\mathrm{thanks}\mathrm{a}\mathrm{lot}.\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{first}\mathrm{time}\mathrm{i}\mathrm{see}$$$$\mathrm{this}\mathrm{method}.\mathrm{ill}\mathrm{get}\mathrm{it}\mathrm{to}\mathrm{very}\mathrm{good}\mathrm{use}$$$$\mathrm{next}\mathrm{time}$$

Commented by Frix last updated on 09/Aug/24

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{important}\mathrm{to}\mathrm{know}\mathrm{when}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}$$$$\mathrm{to}\mathrm{use}\mathrm{and}\mathrm{when}\mathrm{not}.$$

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