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Question-14667




Question Number 14667 by Umar math last updated on 03/Jun/17
Commented by prakash jain last updated on 03/Jun/17
(1/((x^3 −1)^3 ))=(1/((x−1)^3 (x^2 +x+1)^3 ))  =(A/(x−1))+(B/((x−1)^2 ))+(C/((x−1)^3 ))  =((Dx+E)/(x^2 +x+1))+((Fx+G)/((x^2 +x+1)^2 ))+((Hx+I)/((x^2 +x+1)^3 ))  computing  (5/(27(x−1)))−(1/(9(x−1)^2 ))+(1/(27(x−1)^3 ))  +((−2x−1)/(9(x^2 +x+1)^3 ))+((−7x−8)/(27(x^2 +x+1)^2 ))+((−5x−7)/(27(x^2 +x+1)))  Each of the term in the partial  fraction can be integrated easily.  However this is a lengthy procedure.
$$\frac{\mathrm{1}}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{C}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{{Dx}+{E}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{{Fx}+{G}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{Hx}+{I}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${computing} \\ $$$$\frac{\mathrm{5}}{\mathrm{27}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{9}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{27}\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$+\frac{−\mathrm{2}{x}−\mathrm{1}}{\mathrm{9}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{−\mathrm{7}{x}−\mathrm{8}}{\mathrm{27}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{−\mathrm{5}{x}−\mathrm{7}}{\mathrm{27}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{partial} \\ $$$$\mathrm{fraction}\:\mathrm{can}\:\mathrm{be}\:\mathrm{integrated}\:\mathrm{easily}. \\ $$$$\mathrm{However}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{lengthy}\:\mathrm{procedure}. \\ $$
Commented by Umar math last updated on 04/Jul/17
thanks sir.
$${thanks}\:{sir}. \\ $$
Answered by Umar math last updated on 04/Jun/17
thank u sir
$${thank}\:{u}\:{sir} \\ $$

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