5. Let $a_{i} \in \mathbf{R}(i=1,2, \cdots, 10)$, and $\sum_{i=1}^{10} \frac{a_{i}^{2}}{a_{i}^{2}+1}=1$.
Then the range of $\sum_{i=1}^{10} \frac{a_{i}}{a_{i}^{2}+1}$ is
5. $[-3,3]$.
$$
\text { Given } \sum_{i=1}^{10} \frac{a_{i}^{2}}{a_{i}^{2}+1}=1 \Rightarrow \sum_{i=1}^{10} \frac{1}{a_{i}^{2}+1}=9 \text {. }
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
9=\left(\sum_{i=1}^{10} \frac{a_{i}^{2}}{a_{i}^{2}+1}\right)\left(\sum_{i=1}^{10} \frac{1}{a_{i}^{2}+1}\right) \\
\geqslant\left(\sum_{i=1}^{10} \frac{a_{i}}{a_{i}^{2}+1}\right)^{2} \\
\Rightarrow-3 \leqslant \sum_{i=1}^{10} \frac{a_{i}}{a_{i}^{2}+1} \leqslant 3 .
\end{array}
$$

Therefore, the range of $\sum_{i=1}^{10} \frac{a_{i}}{a_{i}^{2}+1}$ is $[-3,3]$.