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Question Number 43731 by Meritguide1234 last updated on 14/Sep/18

	Answered by MJS last updated on 14/Sep/18
	$∫(((x^2 −3x+(1/3))/(x^3 −x+1)))^2 dx=(1/9)∫(((3x^2 +9x+1)^2 )/((x^3 −x+1)^2 ))dx once again my beloved method of Ostrogradski ∫((P(x))/(Q(x)))dx=((P_1 (x))/(Q_1 (x)))+∫((P_2 (x))/(Q_2 (x)))dx Q_1 (x)=gcf(Q(x), Q^′ (x))=x^3 −x+1 Q_2 (x)=((Q(x))/(Q_1 (x)))=x^3 −x+1 P_1 (x)=ax^2 +bx+c P_2 (x)=dx^2 +ex+f the factors can be found by setting ((P(x))/(Q(x)))=((P_1 ′(x)Q_1 (x)−P_1 (x)Q_1 ′(x))/(Q_1 ^2 (x{))+((P_2 (x))/(Q_2 (x))) P_1 (x)=−9x^2 +27x−26 P_2 (x)=0 ⇒ (1/9)∫(((3x^2 +9x+1)^2 )/((x^3 −x+1)^2 ))dx=−((9x^2 −27x+26)/(9(x^3 −x+1)))+C$
	$\int {(\frac{x^{2} - 3 x + \frac{1}{3}}{x^{3} - x + 1})}^{2} d x = \frac{1}{9} \int \frac{{(3 x^{2} + 9 x + 1)}^{2}}{{(x^{3} - x + 1)}^{2}} d x$ $once again my beloved method of Ostrogradski$ $\int \frac{P (x)}{Q (x)} d x = \frac{P_{1} (x)}{Q_{1} (x)} + \int \frac{P_{2} (x)}{Q_{2} (x)} d x$ $Q_{1} (x) = gcf (Q (x), Q^{^{'}} (x)) = x^{3} - x + 1$ $Q_{2} (x) = \frac{Q (x)}{Q_{1} (x)} = x^{3} - x + 1$ $P_{1} (x) = {a x}^{2} + b x + c$ $P_{2} (x) = {d x}^{2} + e x + f$ $the factors can be found by setting$ $\frac{P (x)}{Q (x)} = \frac{P_{1}^{'} (x) Q_{1} (x) - P_{1} (x) Q_{1}^{'} (x)}{Q_{1}^{2} (x {} + \frac{P_{2} (x)}{Q_{2} (x)}$ $P_{1} (x) = - 9 x^{2} + 27 x - 26$ $P_{2} (x) = 0$ $\Rightarrow \frac{1}{9} \int \frac{{(3 x^{2} + 9 x + 1)}^{2}}{{(x^{3} - x + 1)}^{2}} d x = - \frac{9 x^{2} - 27 x + 26}{9 (x^{3} - x + 1)} + C$
	Commented by Meritguide1234 last updated on 15/Sep/18

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