I don't know of any easier examples, but here is one. Let $k$ be a field and $R$ be the set equivalence classes of elements of the form $\frac{f(x,y)}{g(x,y)}$ where $x,y$ are indeterminates over $k$, $f,g\in k[x,y]$, $x$ does not divide $g$ (in $k[x,y]$) and $\frac{f(0,y)}{g(0,y)}\in k$. Make $R$ into a ring by the usual addition and multiplication of rational functions. Then, it can be shown that $R$ is a commutative ring that is not noetherian and every ideal of $R$ has a primary decomposition.
This is an example from Gilmer's paper linked here.
He also gives a characterization of rings in which every ideal has a unique primary decomposition.
Ok, here's a sketch.
First note that $\varinjlim R[x_1,x_2,\ldots,x_j]=R[x_1,x_2,\ldots,]$. Also note that the direct limit is an exact functor (see the exercises of Atiyah-MacDonald for example).
Let $M$ be a finitely generated ideal of $R[x_1,x_2,\ldots]$.
It is generated by some $f_1,\ldots,f_n$, which all lie in some $R[x_1,\ldots,x_N]$ for some big $N$.
As an ideal in $R[x_1,\ldots,x_N]$, $\langle f_1,\ldots,f_n \rangle$ is finitely presented. Thus we have an exact sequence
$$
R[x_1,\ldots,x_N]^m \to R[x_1,\ldots,x_N]^n \to \langle f_1,\ldots,f_n \rangle \to 0.
$$
of $R[x_1,\ldots,x_N]$-modules.
Now apply $\varinjlim (-)$ to get an exact sequence
$$
\varinjlim R[x_1,\ldots,x_N]^m \to \varinjlim R[x_1,\ldots,x_N]^n \to \varinjlim \langle f_1,\ldots,f_n \rangle \to 0.
$$
which is equal to
$$
R[x_1,x_2,\ldots]^m \to R[x_1,x_2,\ldots]^n \to \langle f_1,\ldots,f_n \rangle=M \to 0
$$
as $R[x_1,x_2,\ldots,]$-modules. The above exact sequence says that $M$ is finitely presented.
Best Answer
In the paper The Laskerian Property, Power Series Rings and Noetherian Spectra, Gilmer and Heinzer show that a Laskerian ring has Noetherian spectrum. But a polynomial ring in countable many indeterminates over a field has no Noetherian spectrum, so it can't be Laskerian.