The usual proofs work in the holomorphic category directly. No need for almost complex structures. Take a holomorphic subbundle $V \subset TM$ of the tangent bundle of a complex manifold $M$. In local holomorphic coordinates $x,y$ we can arrange by linear algebra at some point, say at the origin, that $V_{(0,0)}$ is $dy=A \, dx$ for some matrix $A$. Then we will see that nearby, $V=dy-f(x,y)dx$, some $f(x,y)$ holomorphic. Consider the map $\phi(x,y,v,w)=(x,y,v)$ for $x,v \in \mathbb{C}^p$ and $y,w \in \mathbb{C}^q$. This map is a submersion $\phi \colon TM \to M \times \mathbb{C}^p$. Restricted to $V$, it is (by dimension count) a local biholomorphism, so has an inverse, $w=f(x,y)v$, linear in $v$ since the fibers $V_{(x,y)}$ are linear. Write this as saying that $V=(dy=f(x,y)dx)$. Take each holomorphic vector field $X(x)$ in the $x$-plane, and lift it to a unique holomorphic vector field $\hat{X}(x,y)$ which projects via $(x,y) \mapsto x$ to $X$. This holomorphic vector field is $\hat{X}(x,y)=(X(x),f(x,y)X(x))$, clearly by linear algebra. Because the holomorphic lifts project to the origin vector fields, their flows project to their flows, so their brackets project to their brackets. So if you lift commuting vector fields, like constant ones, you get ones with vertical brackets. If $V$ is bracket closed, you get zero brackets of those lifts. So commuting vector fields lift to commuting vector fields. The flows of the lifts then give holomorphic integral manifolds, by the holomorphic implicit function theorem. There is a unique one through each point $(x,y)=(0,y)$, and then the flow times along the fields together with the initial $y$ value make holomorphic coordinates in which the fields are coordinates translations.

The usual proofs work in the holomorphic category directly. No need for almost complex structures. Take a holomorphic subbundle $V \subset TM$ of the tangent bundle of a complex manifold $M$. In local holomorphic coordinates $x,y$ we can arrange by linear algebra at some point, say at the origin, that $V_{(0,0)}$ is $dy=A \, dx$ for some matrix $A$. Then we will see that nearby, $V=(dy-f(x,y)dx)$, some $f(x,y)$ holomorphic. Consider the map $\phi(x,y,v,w)=(x,y,v)$ for $x,v \in \mathbb{C}^p$ and $y,w \in \mathbb{C}^q$. This map is a submersion $\phi \colon TM \to M \times \mathbb{C}^p$. Restricted to $V$, it is (by dimension count) a local biholomorphism, so has an inverse, $w=f(x,y)v$, linear in $v$ since the fibers $V_{(x,y)}$ are linear. Take each holomorphic vector field $X(x)$ in the $x$-plane, and lift it to a unique holomorphic vector field $\hat{X}(x,y)$ which projects via $(x,y) \mapsto x$ to $X$. This holomorphic vector field is $\hat{X}(x,y)=(X(x),f(x,y)X(x))$, clearly by linear algebra. Because the holomorphic lifts project to the origin vector fields, their flows project to their flows, so their brackets project to their brackets. So if you lift commuting vector fields, like constant ones, you get ones with vertical brackets. If $V$ is bracket closed, you get zero brackets of those lifts. So commuting vector fields lift to commuting vector fields. The flows of the lifts then give holomorphic integral manifolds, by the holomorphic implicit function theorem. There is a unique one through each point $(x,y)=(0,y)$, and then the flow times along the fields together with the initial $y$ value make holomorphic coordinates in which the fields are coordinates translations.

The usual proofs work in the holomorphic category directly. No need for almost complex structures. Take a holomorphic subbundle $V \subset TM$ of the tangent bundle of a complex manifold $M$. In local holomorphic coordinates $x,y$ we can arrange by linear algebra at some point, say at the origin, that $V$ is $dy=0$. Then nearby, $V=dy-f(x,y)dx$, some $f(x,y)$ holomorphic, by the holomorphic implicit function theorem. Take each holomorphic vector field $X(x)$ in the $x$-plane, and lift it to a unique holomorphic vector field $\hat{X}(x,y)$ which projects via $(x,y) \mapsto x$ to $X$. Because the holomorphic lifts project to the origin vector fields, their flows project to their flows, so their brackets project to their brackets. So if you lift commuting vector fields, like constant ones, you get ones with vertical brackets. If $V$ is bracket closed, you get zero brackets of those lifts. So commuting vector fields lift to commuting vector fields. The flows of the lifts then give holomorphic integral manifolds, by the holomorphic implicit function theorem. There is a unique one through each point $(x,y)=(0,y)$, and then the flow times along the fields together with the initial $y$ value make holomorphic coordinates in which the fields are coordinates translations.

The usual proofs work in the holomorphic category directly. No need for almost complex structures. Take a holomorphic subbundle $V \subset TM$ of the tangent bundle of a complex manifold $M$. In local holomorphic coordinates $x,y$ we can arrange by linear algebra at some point, say at the origin, that $V_{(0,0)}$ is $dy=A \, dx$ for some matrix $A$. Then we will see that nearby, $V=dy-f(x,y)dx$, some $f(x,y)$ holomorphic. Consider the map $\phi(x,y,v,w)=(x,y,v)$ for $x,v \in \mathbb{C}^p$ and $y,w \in \mathbb{C}^q$. This map is a submersion $\phi \colon TM \to M \times \mathbb{C}^p$. Restricted to $V$, it is (by dimension count) a local biholomorphism, so has an inverse, $w=f(x,y)v$, linear in $v$ since the fibers $V_{(x,y)}$ are linear. Write this as saying that $V=(dy=f(x,y)dx)$. Take each holomorphic vector field $X(x)$ in the $x$-plane, and lift it to a unique holomorphic vector field $\hat{X}(x,y)$ which projects via $(x,y) \mapsto x$ to $X$. This holomorphic vector field is $\hat{X}(x,y)=(X(x),f(x,y)X(x))$, clearly by linear algebra. Because the holomorphic lifts project to the origin vector fields, their flows project to their flows, so their brackets project to their brackets. So if you lift commuting vector fields, like constant ones, you get ones with vertical brackets. If $V$ is bracket closed, you get zero brackets of those lifts. So commuting vector fields lift to commuting vector fields. The flows of the lifts then give holomorphic integral manifolds, by the holomorphic implicit function theorem. There is a unique one through each point $(x,y)=(0,y)$, and then the flow times along the fields together with the initial $y$ value make holomorphic coordinates in which the fields are coordinates translations.