Help me evaluate $\int \arccos(\frac{x^3-3x}{2})dx$ I need to solve $\int \arccos(\frac{x^3-3x}{2})dx$. I tried integration by parts by adding an $x'$, but it didn't work. I also tried a change of variable with $\cos(t) = \frac{x^3-3x}{2}$, but that didn't get anywhere either. Could you point me in the right direction?

integrate by parts: Let $f=\arccos\left(\frac{x^3 - 3x}{2}\right)$, $g'=1$. Then $g=x$ and $$f'= -\frac{3(x^2-1)}{2\sqrt{1-\frac{1}{4}(x^3-3x)^2}}.$$ Thus 
$$\begin{align*}
\int fg'~dx & = x\arccos\left(\frac{x^3 - 3x}{2}\right)  + \int\frac{3x^3-3x}{2\sqrt{1-\frac{1}{4}(x^3-3x)^2}}~dx \\
&= x\arccos\left(\frac{x^3 - 3x}{2}\right) + \int\frac{3x(x^2-1)}{\sqrt{4-(x^3-3x)^2}}~dx \\

\end{align*}$$  
At this point you note that $$4-(x^3-3x)^2 = -(x^2-1)^2(x^2-4).$$
Can you take if from here?
