Here's a general integration-by-parts using Stokes' theorem. I saw this in class, so I don't recall reading this in a text, but I'm sure it's covered in any text that covers differential forms --- try Principles of Mathematical Analysis (Rudin), or Differential Topology (Guillemin-Pollack).
Remember that if $\alpha,\beta$ are respectively $k-1$ and $n-k$-forms on $M$, then $d(\alpha\wedge\beta) = (d\alpha)\wedge\beta + (-1)^k\alpha\wedge (d\beta).$ We have required that $\alpha$ and $\beta$ be $k-1$ and $n-k$ forms so that $d(\alpha\wedge\beta)$ is a multiple of the volume form and so can be integrated against $M$ instead of a submanifold of $M$.
Now we can use this formula and Stokes' theorem: \begin{align*}\int_M d\alpha\wedge\beta &= \int_M \bigg(d(\alpha\wedge\beta) - (1)^k\alpha\wedge (d\beta)\bigg) \\ &= \int_{\partial M}\alpha\wedge\beta + (-1)^{k+1} \int_M \alpha\wedge (d\beta) \end{align*} Note that if $M$ is closed, the term integrating over the boundary vanishes.
This formula recovers integration by parts on the real line. Let $M = [a,b]\subset \mathbb{R}$, so that $\partial M = \{a\}\cup\{b\}$. If $f$ is a function, then $df = \frac{df}{dx}dx$. Integrating by parts is expressed by: $\int_a^b \frac{df}{dx}gdx = \int_a^b df\wedge g = \int_{\partial M} fg - \int_a^b f\wedge dg = (fg)\bigg|_a^b - \int_a^bf\frac{dg}{dx}dx.$ Integration of the zero-form $fg$ over $\partial M$ is signed evaluation.
In response to the question in comments below, integrating a top-dimensional form is exactly the same as integrating a function against the measure induced by the volume form. On an orientable manifold $M$, the top exterior bundle $\Lambda^nM$ is a trivializable rank $1$ bundle. Choice of a volume form $Vol$ trivializes the bundle, hence induces an isomorphism ($C^\infty(M)$-modules) $\Omega^n(M)\cong C^\infty(M)$, where $1\leftrightarrow Vol$. So integrating a function against the volume measure is the same as integrating its corresponding $n$-form.