How we can compute this integral $\int \frac{1}{(1+\arccos(x))(1-x^2)} dx$?

Question

$$\int \frac{1}{(1+\arccos(x))(1-x^2)} dx$$

i tired first to use substitution $u=\arccos(x)$ then $du=-\frac{1}{\sqrt{1-x^2}}$

but this may not give me what im looking for

as result is there any shortcut to attack this problem >?

Answer 1

I think that only a series solution could work. Use for example $$\frac{1}{1+\cos ^{-1}(x)}=$$ $$\frac{2}{2+\pi }\left(1+\frac{2 x}{2+\pi }+\frac{4 x^2}{(2+\pi )^2}+\frac{(28+4\pi +\pi^2 ) x^3}{3 (2+\pi )^3}+\frac{4 \left(16+4 \pi +\pi ^2\right) x^4}{3 (2+\pi)^4}+O\left(x^5\right) \right)$$ to face integrals looking like $$I_n=\int \frac {x^n}{1-x^2}\,dx$$ which are quite simple after partial fraction decomposition.

Edit

ANothe solution could be to let $x=\cos(u)$ $$\int \frac{dx}{(1+\arccos(x))(1-x^2)} dx=\int \frac{\csc (u)}{u+1} \,du$$ $$\frac{\csc (u)}{u+1}=\sum_{n=0}^\infty (-1)^{n+1}\frac{\left(4^n-2\right) B_{2 n}}{(2 n)!}\frac {u^{2n-1}}{1+u}$$

$$\int \frac {u^{2n-1}}{1+u}=\frac{u^{2n}}{2n}\, _2F_1(1,2 n;2 n+1;-u)$$