5. Given $z_{1}, z_{2}, \cdots, z_{5}$ are the five complex roots of the equation
$$
x^{5}+2 x^{4}+3 x^{3}+4 x^{2}+5 x+6=0
$$

then $\sum_{i=1}^{5} \frac{z_{i}}{z_{i}^{2}+1}=$ $\qquad$
5. $\frac{4}{17}$.

Notice,
$$
\begin{array}{l}
\sum_{i=1}^{5} \frac{z_{i}}{z_{i}^{2}+1}=\frac{1}{2} \sum_{i=1}^{5} \frac{2 z_{i}}{z_{i}^{2}+1} \\
=\frac{1}{2} \sum_{j=1}^{5} \frac{1}{z_{j}+\mathrm{i}}+\frac{1}{2} \sum_{j=1}^{5} \frac{1}{z_{j}-\mathrm{i}}. \\
\text { Let } f(x)=x^{5}+2 x^{4}+3 x^{3}+4 x^{2}+5 x+6=0. \text { Then } \\
\sum_{j=1}^{5} \frac{1}{z_{j}-t}=\frac{f^{\prime}(t)}{f(t)} \\
=\frac{5 t^{4}+8 t^{3}+9 t^{2}+8 t+5}{t^{5}+2 t^{4}+3 t^{3}+4 t^{2}+5 t+6} \\
\Rightarrow \sum_{j=1}^{5} \frac{1}{z_{j}-\mathrm{i}}=\frac{1}{-\mathrm{i}+2-3 \mathrm{i}-4+5 \mathrm{i}+6}=\frac{1}{4+\mathrm{i}} \\
\Rightarrow \sum_{j=1}^{5} \frac{1}{z_{j}+\mathrm{i}}=\frac{1}{\mathrm{i}+2+3 \mathrm{i}-4-5 \mathrm{i}+6}=\frac{1}{4-\mathrm{i}}. \\
\text { Therefore, } \sum_{i=1}^{5} \frac{z_{i}}{z_{i}^{2}+1}=\frac{1}{2} \sum_{j=1}^{5} \frac{1}{z_{j}+\mathrm{i}}+\frac{1}{2} \sum_{j=1}^{5} \frac{1}{z_{j}-\mathrm{i}} \\
=\frac{1}{2}\left(\frac{1}{4-\mathrm{i}}+\frac{1}{4+\mathrm{i}}\right)=\frac{4}{17}. \\
\end{array}
$$