Question-43731

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

Answered by MJS last updated on 14/Sep/18

\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