There is numerical evidence that
$$I=\int_0^1\frac{1}{\sqrt{1-x^2}}\arccos\left(\frac{3x^3-3x+4x^2\sqrt{2-x^2}}{5x^2-1}\right)\mathrm dx=\frac{3\pi^2}{8}-2\pi\arctan\frac12.$$
How can this be proved?
Wolfram does not find an antiderivative.
Here is the graph of $y=\frac{1}{\sqrt{1-x^2}}\arccos\left(\frac{3x^3-3x+4x^2\sqrt{2-x^2}}{5x^2-1}\right)$.
Based on recent experience with integrals involving inverse trigonometric functions (example), I guess a proof may involve a lot of substitutions. But I don't have any insight on how to approach this.
A search on approachzero did not turn up anything similar.
Context
If this can be proved, then we can answer the question Probability that the centroid of a triangle is inside its incircle, via @user170231's answer.
How I found the conjectured closed form
In my comment to @user170231's answer to the linked question, I note that the probability in that question is $1-\frac{12}{\pi^2}\int_0^{\pi/2}(y-x_+)dy$, which is equivalent to $1-\frac{12}{\pi^2}(\int_0^{\pi/2}ydy-I)\approx0.457993176$, where $I$ is the integral in this question.
Wolfram suggests that $0.457993176$ is $4-24\left(\frac{\arctan(1/2)}{\pi}\right)$, which implies that $I=\frac{3\pi^2}{8}-2\pi\arctan\frac12$.
