Answer
$$\int 4 \arccos x d x =4x\arccos x -4\sqrt{1-x^{2}}+c$$
Work Step by Step
Since $$
\int 4 \arccos x d x
$$
integrate by parts , let
\begin{align*}
u&=\arccos (x) \ \ \ \ dv=4 d x \\
du&=-\frac{1}{\sqrt{1-x^{2}}} \ \ \ \ \ v=4 x\end{align*}
Then
\begin{align*}
\int 4 \arccos x d x&=uv-\int vdu \\
&=4x\arccos x +\int \frac{4xdx}{\sqrt{1-x^{2}}}\\
&=4x\arccos x -4\sqrt{1-x^{2}}+c
\end{align*}
