\( \frac {1}{(x^2+1)(x^4-x^2+1)}\)
\( \frac{A}{x^2+1} + \frac{Bx^2 + C}{x^4-x^2+1}\)
\( (A)(x^4-x^2+1) + (Bx^2+C)(x^2+1) = 1\)
\((x^0): A + C = 1 \Rightarrow C = 1 - A\)
\((x^4): A + B = 0 \Rightarrow B = -A\)
\((x^2): -A + B + C = 0 \Rightarrow -A - A + 1 - A =0 \Rightarrow -3A = -1\)
\( A = \frac {1}{3} ; B = \frac {-1}{3} ; C = \frac {2}{3}\)
\(\frac{1}{3(x^2+1)} + \frac{\frac{-x^2}{3} + \frac{2}{3}}{x^4-x^2+1}\)
\( = \frac{1}{3(x^2+1)} + \frac {-(x^2 -2)}{3(x^4-x^2+1)}\)
\( = \frac {\arctan(x)}{3} - \frac {1}{3} \int \frac{x^2 - 2}{x^4-x^2+1} dx\)
Nu zie ik wel niet echt hoe het hierna verder gaat ...