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

Commented by jirodomo last updated on 08/Aug/24

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

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

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

Where′s the issue? x^3 −1=(x−1)(x^2 +x+1) ⇒ ∫((3x^5 −x^4 +2x^3 −12x^2 −2x+1)/((x^3 −1)^2 ))dx= =∫(((2(2x+1))/((x^2 +x+1)^2 ))+((2x+1)/(x^2 +x+1))−(1/((x−1)^2 ))+(1/(x−1)))dx= ... =((x^2 −x+3)/(x^3 −1))+ln ∣x^3 −1∣ +C

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

Commented by jirodomo last updated on 08/Aug/24

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

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

Commented by Frix last updated on 08/Aug/24

Search the www for Ostrogradski′s Method to see the proof. ∫((3x^5 −x^4 +2x^3 −12x^2 −2x+1)/((x^3 −1)^2 ))dx= =((Ax^2 +Bx+C)/(x^3 −1))+∫((Dx^2 +Ex+F)/(x^3 −1))dx Now differentiate both sides: ((3x^5 −x^4 +2x^3 −12x^2 −2x+1)/((x^3 −1)^2 ))= =−((Ax^4 +2Bx^3 +3Cx^2 +2Ax+B)/((x^3 −1)^2 ))+ +((Dx^2 +Ex+F)/(x^3 −1)) Multiplicate with (x^3 −1)^2 and 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 ⇒A=1, B=−1, C=D=3, E=F=0 Now we have ∫((3x^5 −x^4 +2x^3 −12x^2 −2x+1)/((x^3 −1)^2 ))dx= =((x^2 −x+3)/(x^3 −1))+3∫(x^2 /(x^3 −1))dx

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

Commented by jirodomo last updated on 09/Aug/24

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

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

Commented by Frix last updated on 09/Aug/24

It′s important to know when it′s possible to use and when not.

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

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