Ans: B
$\begin{array}{cl}
& \dfrac{i}{k-i}+\dfrac{2}{k+i} \\
= & \dfrac{i(k+i)+2(k-i)}{(k-i)(k+i)} \\
= & \dfrac{ik+i^2+2k-2i}{k^2-i^2} \\
= & \dfrac{ik-1+2k-2i}{k^2-(-1)} \\
= & \dfrac{2k-1+(k-2)i}{k^2+1} \\
= & \dfrac{2k-1}{k^2+1}+\dfrac{k-2}{k^2+1}i
\end{array}$
Therefore, the real part is $\dfrac{2k-1}{k^2+1}$.
