Let $A$ be a (not necessarily commutative, associative) $k$-algebra. The bimodule of non-commutative one-forms $\Omega^1_A$ is the free $A$-bimodule generated by symbols $da$, $a \in A$, subject to the relations $$ d(\lambda a + \mu b) = \lambda da + \mu db, ~~~~~~ d(ab) = a\,db + da \,b$$ for $a, b \in A$, $\lambda, \mu \in k$. Higher degree forms are then defined by $$\Omega^n_A = \Omega^1_A \otimes_A \cdots \otimes_A \Omega^1_A.$$ The direct sum of all forms $\Omega_A = \bigoplus_{n=0}^\infty \Omega_A^n$ becomes a differential graded algebra with the obvious differential.

Now Loday's book "cyclic cohomology", he defines the non-commutative de-Rham cohomology of $A$ to be the cohomology of the complex $((\Omega_A)_{\mathrm{ab}}, d)$, i.e. he takes the abelianization of $\Omega_A$.

**Q:** Why is that? What is wrong with defining the non-commutative de-Rham-cohomology just as the cohomology of $\Omega_A$?

/edit: To answer the question regarding the definition of $d$: Using the relations of $\Omega^1_A$, every element of $\Omega^n_A$ can be brought into the form $a_0 da_1\cdots da_n$. For these, $$d(a_0 da_1\cdots da_n) = da_0da_1 \cdots da_n.$$