Compact support of differential forms is not a necessary condition for the integrals to make sense; but it is definitely a sufficient condition. If you want to be fully general, then the condition is that the differential form be in $L^1$ (one has to define this carefully). To make sense of any type of integral (Lebesgue, or Riemann, or improper Riemann) there's two things to keep track of: regularity of the objects involved (functions/forms/scalar densities you wish to integrate and the sets over which you want to integrate) and the finiteness of stuff involved.
As a special case, let $U\subset\Bbb{R}^n$ be an open set and $\omega$ an $n$-form on $U$ (a-priori, we don't have to make any continuity/smoothness assumptions). Then, we can write $\omega=g\,dx^1\wedge\cdots\wedge dx^n$ for some unique function $g:U\to\Bbb{R}$, where $(x^1,\dots, x^n)$ are the standard cartesian coordinates on $U$. We say that the $n$-form $\omega$ is (Lebesgue) integrable over $U$ if $g\in L^1(U)$, and in this case, we define $\int_U\omega:=\int_Ug\,d\lambda\equiv \int_Ug$, where the integral on the right is the standard Lebesgue integral of $g$ over the open set $U$.
If you only want to consider Riemann integrals then the definition becomes slightly tricky to state, because first of all Riemann integrals are only properly defined for closed rectangles and bounded functions. Then, one has to jump through several hoops to define, say, integrals of locally bounded, a.e continuous functions (which is certainly the case for continuous or smooth functions/forms) on an arbitrary open set $U$ (possibly unbounded). These details can be found in Spivak's Calculus on Manifolds, in the section on partitions of unity (which is somewhere in Chapter 3- Integration). Note that in the sense that Spivak defines things, we still have that $g$ is (Riemann) integrable over $U$ (in the extended sense) if and only if $|g|$ is; so this is very much analogous to the Lebesgue theory.
Note that with the correct definitions, you can certainly consider differential forms such as $\omega=\frac{1}{1+t^2}\,dt$ on $\Bbb{R}$, or $\eta=\frac{1}{\sqrt{t}}\,dt$ on $(0,1)$. These forms are smooth, do not have compact support, but they are still integrable in either of the senses described above (and $\int_{\Bbb{R}}\omega=\pi$ and $\int_{(0,1)}\eta=2$). But note that we need some sort of restrictions, because as you're hopefully familiar with from basic calculus, we have $\int_0^1\frac{dt}{t}=\infty$, so even with smooth functions/forms on a bounded open set, the result might still integrate to $\infty$.
Full Definition.
For the sake of completeness, I'll write out the definition in the fully general case. However, there are a few details to be verified to ensure that this stuff is well-defined; I'll omit these details.
Definition.
Let $M$ be a smooth oriented manifold (assumed Hausdorff, second-countable), and $\omega$ a rough-differential form on $M$. We say $\omega$ is (Lebesgue) integrable on $M$ if the following conditions are satisfied:
- Let $\{(U_i,\phi_i)\}_{i=1}^{N}$, where $1\leq N\leq \infty$, be an at most-countable positively-oriented atlas of $M$, and let $\{\pi_i\}_{i=1}^N$ be a continuous partition of unity of $M$, subordinate to the open cover $\{U_i\}_{i=1}^N$.
- Suppose $\omega$ is Lebesgue measurable, meaning that for chart $(U_i,\phi_i=(x_i^1\cdots, x_i^n))$, if write $\omega|_{U_i}=g_i\cdot dx_i^1\wedge\cdots \wedge dx_i^n$ for some function $g_i:U_i\to\Bbb{R}$, then we require that $g_i\circ \phi_i^{-1}:\phi_i(U_i)\subset\Bbb{R}^n\to\Bbb{R}$ is Lebesgue measurable.
- Finally, we require that $\sum_{i=1}^N\int_{\phi_i(U_i)}|\pi_i\cdot g_i|\circ \phi_i^{-1}\,\,d\lambda<\infty$ (these are standard Lebesgue integrals in open subsets of $\Bbb{R}^n$).
In this case we define $\int_M\omega:= \sum_{i=1}^N\int_{\phi_i(U_i)}(\pi_i\cdot g_i)\circ \phi_i^{-1}\,d\lambda$.
Now, one can easily show that Lebesgue measurability doesn't depend on the choice of atlas $\{(U_i,\phi_i)\}$ (or if you wish to ignore these technicalities, you can assume from the beginning that all forms are smooth). The main thing to be checked is that the third bullet point doesn't depend on the choice of positively oriented atlas and partition of unity $\{(U_i,\phi_i,\pi_i)\}$; here you need Tonelli's theorem and also that you're dealing with an oriented manifold, so that the absolute value of the Jacobian from the change-of-variables theorem can be dropped. Finally, one has to show that the definition of $\int_M\omega$ (which doesn't have absolute values in it) also doesn't depend on the choice $\{(U_i,\phi_i,\pi_i)\}$ (follows by Fubini's theorem/dominated convergence which allows us to swap series with integrals and also change of variables theorem).
Note that if $\omega$ has compact support, then the third bullet point is automatically satisfied since all but finitely many summands vanish. Hence, continuous, compactly supported forms are automatically integrable.
Having said all of this, note that unless you really want to go down the integration rabbit hole, compact support is good enough, because one of the most famous theorems, Stokes theorem, fails (unless you make serious modifications) without this hypothesis:
Stokes Theorem.
Let $M$ be a smooth $n$-dimensional oriented manifold with boundary (possibly empty), let $\omega$ be a smooth $(n-1)$-form on $M$ with compact support. Then, $\int_Md\omega=\int_{\partial M}\omega$.
As a counterexample for forms without compact support, consider $M=(0,1)$, and the function $f(t)=t$ (i.e a $0$-form). Then $M$ is a 1-dimensional smooth manifold, with empty boundary, and $f$ is a smooth 0-form without compact support (it's support equals $M$ which is non-compact), even though both $f$ and $df$ are integrable forms (on $\partial M=\emptyset$ and $M=(0,1)$ respectively). Then,
\begin{align}
\int_Mdf=\int_{(0,1)}dt=1\neq 0=\int_{\emptyset}f=\int_{\partial M}f.
\end{align}
(keep in mind that $\partial M=\partial ((0,1))=\emptyset$; here $\partial$ is talking about the manifold boundary, which doesn't always coincide with the topological boundary $\text{bd}_{\Bbb{R}}(M)=\{0,1\}$ of $M=(0,1)$ as a subset of $\Bbb{R}$).
