Let me try to give a counterexample. (I don't know whether it is 'nice'). First, let us rewrite your properties for an affine scheme $X=Spec(A)$.

Connectedness for $A$ means $A$ has no nontrivial idempotents;

Integrality for $A$ is the usual one ($A$ is a domain);

Local integrality means that whenever $fg=0$ in $A$, every point of $X$ has a neighborhood
where either $f$ or $g$ vanishes.

Let us construct a connected locally integral ring that is not integral.

Roughly speaking, the construction is as follows: let $X_0$ be the cross (the union of coordinate axes) on the affine plane. Then let $X_1$ be the (reduced) full preimage of $X_0$ on the blow-up of the plane ($X_1$ has three rational components forming a chain). Then blow up the resulting surface at the two singularities of $X_1$, and let $X_2$ be the reduced preimage of $X_1$
(which has five rational components), etc. Take $X$ to be the inverse limit.

The only problem with this construction is that blow-ups glue in a projective line, so $X_1$ is not affine. Let us correct this by gluing in an affine line instead (so our scheme will be an open subset in what was described above).

Here's an algebraic description:

For every $k\ge 0$, let $A_k$ be the following ring: its elements are collections of
polynomials $p_i\in{\mathbb C}[x]$ where $i=0,\dots,2^k$ such that $p_i(1)=p_{i+1}(0)$.
Set $X_k=Spec(A_k)$. $X$ is a union of $2^k+1$ affine lines that meet transversally in a chain. (It may be better to index polynomials by $i/2^k$, but the notation gets confusing.)

Define a morphism $A_k\to A_{k+1}$ by
$$(p_0,\dots,p_{2^k})\mapsto(p_0,p_0(1),p_1,p_1(1),\dots,p_{2^k})$$
(every other polynomial is constant). This identifies $A_k$ with a subring of $A_{k+1}$.
Let $A$ be the direct limit of $A_k$ (basically, their union). Set $X=Spec(A)$. For every
$k$, we have a natural embedding $A_k\to A$, that is, a map $X\to X_k$.

Each $A_k$ is connected but not integral; this implies that $A$ is connected but not integral. It remains to show that $A$ is locally integral.

Take $f,g\in A$ with $fg=0$ and $x\in X$. Let us construct a neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$ such that $f,g\in A_{k-1}$ (note the $k-1$ index).
Let $y$ be the image of $x$ on $X_k$. It suffices to prove that $y$ has a neighborhood on
which either $f$ or $g$ (viewed as functions on $X_k$) vanishes.

If $y$ is a smooth point of $X_k$ (that is, it lies on only one of the $2^k+1$ lines), this is obvious. We can therefore assume that $y$ is one of the $2^k$ singular points, so two components of $X_k$ pass through $y$. However, on one of these two components (the one with odd index), both $f$ and $g$ are constant, since they are pullbacks of functions on $X_{k-1}$. Since $fg=0$ everywhere, either $f$ or $g$ (say, $f$) vanishes on the other component.
This implies that $f$ vanishes on both components, as required.