5. The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=1, \\
x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}(i=1,2, \cdots) .
\end{array}
$$

Then the general term formula $x_{n}=$ . $\qquad$
5. $\frac{n^{2}+n}{2}$.

From $x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}$, we know $x_{i+1}-x_{i} \geqslant 0$.
Since $x_{1}=1$, we have
$$
x_{i+1} \geqslant x_{i} \geqslant 1(i=1,2, \cdots) \text {. }
$$

Squaring both sides of $x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}$ and rearranging, we get
$$
x_{i+1}^{2}-\left(2 x_{i}+1\right) x_{i+1}+x_{i}^{2}-x_{i}=0 .
$$

By the quadratic formula, we have
$$
x_{i+1}=\frac{2 x_{i}+1 \pm \sqrt{8 x_{i}+1}}{2} \text {. }
$$

If $x_{i+1}=\frac{2 x_{i}+1-\sqrt{8 x_{i}+1}}{2}$, then
$$
x_{i+1}-x_{i}=\frac{1-\sqrt{8 x_{i}+1}}{2}<0 \text {, }
$$

which contradicts equation (1).
$$
\begin{array}{l}
\text { Hence } x_{i+1}=\frac{2 x_{i}+1+\sqrt{8 x_{i}+1}}{2} \\
\Rightarrow 8 x_{i+1}=8 x_{i}+4 \sqrt{8 x_{i}+1}+4 \\
\Rightarrow 8 x_{i+1}+1=\left(\sqrt{8 x_{i}+1}+2\right)^{2} \\
\Rightarrow \sqrt{8 x_{i+1}+1}=\sqrt{8 x_{i}+1}+2 \\
\Rightarrow \sqrt{8 x_{n}+1}=3+2(n-1) \\
\Rightarrow x_{n}=\frac{n^{2}+n}{2} .
\end{array}
$$