You are correct. Use the partial fraction decomposition: http://en.wikipedia.org/wiki/Partial_fraction For example, if $n=4$, the decomposition is (over the rationals):
$$\begin{array}{l} {\frac {ck}{ \left( {k}^{2}+c \right) \left( -c+a \right) \left( -c+ b \right) \left( -d+c \right) }}-\\ {\frac {dk}{ \left( {k}^{2}+d \right) \left( -d+a \right) \left( -d+b \right) \left( -d+c \right) }}-\\ {\frac {bk}{ \left( {k}^{2}+b \right) \left( -b+a \right) \left( -c+b \right) \left( -d+b \right) }}+\\ {\frac {ak}{ \left( {k}^{2}+a \right) \left( -b+a \right) \left( -c+a \right) \left( -d+a \right) }}\end{array} $$ where $a=x_1^2,b=x_2^2,...$. I guess you can get the pattern. For odd $n$ there is no $k$ in the numerator. This is the cause of the difference you noticed.
