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Question Number 14667 by Umar math last updated on 03/Jun/17

Commented by prakash jain last updated on 03/Jun/17
$(1/((x^3 −1)^3 ))=(1/((x−1)^3 (x^2 +x+1)^3 )) =(A/(x−1))+(B/((x−1)^2 ))+(C/((x−1)^3 )) =((Dx+E)/(x^2 +x+1))+((Fx+G)/((x^2 +x+1)^2 ))+((Hx+I)/((x^2 +x+1)^3 )) computing (5/(27(x−1)))−(1/(9(x−1)^2 ))+(1/(27(x−1)^3 )) +((−2x−1)/(9(x^2 +x+1)^3 ))+((−7x−8)/(27(x^2 +x+1)^2 ))+((−5x−7)/(27(x^2 +x+1))) Each of the term in the partial fraction can be integrated easily. However this is a lengthy procedure.$
$\frac{1}{{(x^{3} - 1)}^{3}} = \frac{1}{{(x - 1)}^{3} {(x^{2} + x + 1)}^{3}}$ $= \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}}$ $= \frac{D x + E}{x^{2} + x + 1} + \frac{F x + G}{{(x^{2} + x + 1)}^{2}} + \frac{H x + I}{{(x^{2} + x + 1)}^{3}}$ $c o m p u t i n g$ $\frac{5}{27 (x - 1)} - \frac{1}{9 {(x - 1)}^{2}} + \frac{1}{27 {(x - 1)}^{3}}$ $+ \frac{- 2 x - 1}{9 {(x^{2} + x + 1)}^{3}} + \frac{- 7 x - 8}{27 {(x^{2} + x + 1)}^{2}} + \frac{- 5 x - 7}{27 (x^{2} + x + 1)}$ $Each of the term in the partial$ $fraction can be integrated easily .$ $However this is a lengthy procedure .$
Commented by Umar math last updated on 04/Jul/17

$t h a n k s s i r .$
	Answered by Umar math last updated on 04/Jun/17

	$t h a n k u s i r$
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