What we are trying to find is:
\[ \int \arccos(x) \ dx \]
We are going to use integration by parts:
\[ \int u \ dv = uv - \int v \ du\]
Let's define \(u\) and \(v\):
\[\begin{gathered} u = \arccos(x), du= -\frac{dx}{\sqrt{1-x^2}} \\ v = x , dv = dx \end{gathered}\]
This means:
\[ \int \arccos(x) \ dx = x(\arccos(x)) - \int - \frac{x}{\sqrt{1-x^2}} \ dx\]
Derivating \(\sqrt{1-x^2}\) gives \(-\frac{x}{\sqrt{1-x^2}}\), this means:
\[ \int \arccos(x) \ dx = x(\arccos(x)) - \sqrt{1-x^2} +C\]