How do you solve: 1 / x^2(x^2+2x+2)
I put it in the form:
A/x + B/x^2 + (Cx+D)/(x^2+2x+2) which can be solved by completing the square
A/x + B/x^2 + (Cx+D)/[(x+1)^2 + 1] but now I'm stuck... Please help.
$\displaystyle \frac{1}{x^2(x^2+2x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+2x+2}$
$\displaystyle 1 = Ax(x^2+2x+2) + B(x^2+2x+2) + (Cx+D)x^2$
$\displaystyle 1 = (A+C)x^3 + (2A+B+D)x^2 + (2A+2B)x + 2B$
equating coefficients ...
$\displaystyle A+C = 0$
$\displaystyle 2A+B+D = 0$
$\displaystyle 2A+2B = 0$
$\displaystyle 2B = 1$
$\displaystyle B = \frac{1}{2}$
$\displaystyle A = -\frac{1}{2}$
$\displaystyle C = \frac{1}{2}$
$\displaystyle D = \frac{1}{2}$
$\displaystyle \frac{1}{x^2(x^2+2x+2)} = -\frac{1}{2}\left(\frac{1}{x} - \frac{1}{x^2} - \frac{x+1}{x^2+2x+2}\right)$