$\displaystyle \frac{1+x}{x(1+x^{2})}$
When I expand this, I can't eliminate to get one variable to get the variable.
Write it as
$\displaystyle \frac{A}{x} + \frac{Bx + C}{1 + x^2} = \frac{1 + x}{x(1 + x^2)}$
$\displaystyle \frac{A(1 + x^2) + (Bx + C)x}{x(1 + x^2)} = \frac{1 + x}{x(1 + x^2)}$
$\displaystyle A(1 + x^2) + (Bx + C)x = 1 + x$
$\displaystyle A + Ax^2 + Bx^2 + Cx = 1 + x$
$\displaystyle (A + B)x^2 + Cx + A = 0x^2 + x + 1$.
Equating coefficients of like powers of $\displaystyle x$ gives
$\displaystyle A + B = 0, C = 1, A = 1$.
So $\displaystyle B = -1$.
Therefore:
$\displaystyle \frac{1}{x} + \frac{1 - x}{1 + x^2} = \frac{1 + x}{x(1 + x^2)}$.