∫ ((x^6 −1)/((x^3 −1)^3 )) dx=?


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Question Number 138909 by bramlexs22 last updated on 19/Apr/21

$\int \frac{x^{6} - 1}{{(x^{3} - 1)}^{3}} dx = ?$
	Answered by mathmax by abdo last updated on 20/Apr/21
	$Φ=∫ ((x^6 −1)/((x^3 −1)^3 ))dx⇒Φ=∫ ((x^6 −1)/((x−1)^3 (x^2 +x+1)^3 ))dx the roots of x^2 +x+1=are j=e^((i2π)/3) and j^− =e^(−((i2π)/3)) ⇒ Φ=∫ ((x^6 −1)/((x−1)^3 (x−j)^3 (x−j^− )^3 ))dx =∫ ((x^6 −1)/((x−1)^3 (((x−j)/(x−j^− )))^3 (x−j^− )^6 ))dx we do the changement ((x−j)/(x−j^− ))=t ⇒x−j=tx−j^− t ⇒ (1−t)x=j−j^− t ⇒x=((j−j^− t)/(1−t)) ⇒(dx/dt)=((−j^− (1−t)−(j−j^− t)(−1))/((1−t)^2 )) =((j−j^− )/((1−t)^2 )) =((2i.((√3)/2))/((1−t)^2 )) =((i(√3))/(2(1−t)^2 )) and x−j^− =((j−j^− t)/(1−t))−j^− =((j−j^− t−j^− +j^− t)/(1−t))=((i(√3))/(2(1−t))) also x−1=((j−j^− t)/(1−t))−1 =((j−j^− t−1+t)/(1−t)) =((j−1+(1−j^− )t)/(1−t)) ⇒ Φ =∫ (((((j−j^− t)/(1−t)))^6 −1)/((((j−1+(1−j^− )t)/(1−t)))^3 t^3 (((i(√3))/(2(1−t))))^6 ))((i(√3))/(2(1−t)^2 ))dt =∫ (((((j−j^− t)/(1−t)))^6 −1)(1−t)^9 .2^6 (i(√3)))/((j−1+(1−j^− )t)^3 t^3 .2.(i(√3))^6 (1−t)^2 ))dt =(2^5 /((i(√3))^5 ))∫ (({(((j−j^− t)/(1−t)))^6 −1}(1−t)^7 )/(t^3 (j−1+(1−j^− )t)^3 ))dt =((2/(i(√3))))^5 ∫ (((1−t)(j−j^− t)^6 −(1−t)^7 )/(t^3 (j−1+(1−j^− t))^3 ))dt =((2/(i(√3))))^5 ∫ (((1−t)(j−j^− t)^6 )/(t^3 (j−1+(1−j^− t))^3 ))dt−((2/(i(√3))))^5 ∫ (((1−t)^7 )/(t^3 (j−1+(1−j^− t)^3 ))dt ....be continued....$
	$Φ = \int \frac{x^{6} - 1}{{(x^{3} - 1)}^{3}} dx \Rightarrow Φ = \int \frac{x^{6} - 1}{{(x - 1)}^{3} {(x^{2} + x + 1)}^{3}} dx$ $the roots of x^{2} + x + 1 = are j = e^{\frac{i 2 π}{3}} and \bar{j} = e^{- \frac{i 2 π}{3}} \Rightarrow$ $Φ = \int \frac{x^{6} - 1}{{(x - 1)}^{3} {(x - j)}^{3} {(x - \bar{j})}^{3}} dx = \int \frac{x^{6} - 1}{{(x - 1)}^{3} {(\frac{x - j}{x - \bar{j}})}^{3} {(x - \bar{j})}^{6}} dx$ $we do the changement \frac{x - j}{x - \bar{j}} = t \Rightarrow x - j = tx - \bar{j t} \Rightarrow$ $(1 - t) x = j - \bar{j t} \Rightarrow x = \frac{j - \bar{j t}}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{- \bar{j} (1 - t) - (j - \bar{j t}) (- 1)}{{(1 - t)}^{2}}$ $= \frac{j - \bar{j}}{{(1 - t)}^{2}} = \frac{2 i . \frac{\sqrt{3}}{2}}{{(1 - t)}^{2}} = \frac{i \sqrt{3}}{2 {(1 - t)}^{2}} and x - \bar{j} = \frac{j - \bar{j t}}{1 - t} - \bar{j}$ $= \frac{j - \bar{j t} - \bar{j} + \bar{j t}}{1 - t} = \frac{i \sqrt{3}}{2 (1 - t)} also x - 1 = \frac{j - \bar{j t}}{1 - t} - 1$ $= \frac{j - \bar{j t} - 1 + t}{1 - t} = \frac{j - 1 + (1 - \bar{j}) t}{1 - t} \Rightarrow$ $Φ = \int \frac{{(\frac{j - \bar{j t}}{1 - t})}^{6} - 1}{{(\frac{j - 1 + (1 - \bar{j}) t}{1 - t})}^{3} t^{3} {(\frac{i \sqrt{3}}{2 (1 - t)})}^{6}} \frac{i \sqrt{3}}{2 {(1 - t)}^{2}} dt$ $= \int \frac{{(\frac{j - \bar{j t}}{1 - t})}^{6} - 1) {(1 - t)}^{9} . 2^{6} (i \sqrt{3})}{{(j - 1 + (1 - \bar{j}) t)}^{3} t^{3} . 2 . {(i \sqrt{3})}^{6} {(1 - t)}^{2}} dt$ $= \frac{2^{5}}{{(i \sqrt{3})}^{5}} \int \frac{{{(\frac{j - \bar{j t}}{1 - t})}^{6} - 1} {(1 - t)}^{7}}{t^{3} {(j - 1 + (1 - \bar{j}) t)}^{3}} dt$ $= {(\frac{2}{i \sqrt{3}})}^{5} \int \frac{(1 - t) {(j - \bar{j t})}^{6} - {(1 - t)}^{7}}{t^{3} {(j - 1 + (1 - \bar{j t}))}^{3}} dt$ $= {(\frac{2}{i \sqrt{3}})}^{5} \int \frac{(1 - t) {(j - \bar{j t})}^{6}}{t^{3} {(j - 1 + (1 - \bar{j t}))}^{3}} dt - {(\frac{2}{i \sqrt{3}})}^{5} \int \frac{{(1 - t)}^{7}}{t^{3} (j - 1 + {(1 - \bar{j t})}^{3}} dt$ $. . . . be continued . . . .$
	Answered by phanphuoc last updated on 20/Apr/21

	$\int \frac{x^{3} + 1}{{(x^{3} - 1)}^{2}} d x = \int \frac{d x}{x^{3} - 1} + 2 \int \frac{d x}{{(x^{3} - 1)}^{2}}$
	Answered by MJS_new last updated on 20/Apr/21

	$\int \frac{(x^{3} - 1) (x^{3} + 1)}{{(x^{3} - 1)}^{3}} d x = \int \frac{x^{3} + 1}{{(x^{3} - 1)}^{2}} d x =$ $[{Ostrogradski}^{'} s Method]$ $= - \frac{2 x}{3 (x^{3} - 1)} - \frac{1}{3} \int \frac{d x}{x^{3} - 1} =$ $= - \frac{2 x}{3 (x^{3} - 1)} + \frac{1}{9} \int \frac{x + 2}{x^{2} + x + 1} d x - \frac{1}{9} \int \frac{d x}{x - 1} =$ $= - \frac{2 x}{3 (x^{3} - 1)} + \frac{\sqrt{3}}{9} \arctan \frac{2 x + 1}{\sqrt{3}} + \frac{1}{18} \ln (x^{2} + x + 1) - \frac{1}{9} \ln ∣ x - 1 ∣ + C$
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