[tex]\arccos(-x) = -\arccos x+\pi \\ \\ \text{Aplic formula lui King}:[/tex]
[tex]\displaystyle \boxed{\int_{a}^b f(x)\, dx = \int_{a}^b f(a+b-x)\, dx}[/tex]
[tex]\displaystyle I =\int_{-1}^1 \dfrac{\arccos x}{1+x^2}\, dx =\int_{-1}^1 \dfrac{\arccos(-1+1-x)}{1+(-1+1-x)^2}\, dx = \\ \\ = \int_{-1}^{1} \dfrac{\arccos(-x)}{1+x^2}\, dx = \int_{-1}^{1} \dfrac{-\arccos x+\pi}{1+x^2}\, dx = \\ \\ = -I +\int_{-1}^1 \dfrac{\pi}{1+x^2}\\ \\\\ 2I = \int_{-1}^1 \dfrac{\pi}{1+x^2}\, dx =\\ \\ 2I = \pi\arctan x \Big|_{-1}^1[/tex]
[tex]2I = \dfrac{\pi^2}{4}+\dfrac{\pi^2}{4}\\ \\\\ \Rightarrow \boxed{I = \dfrac{\pi^2}{4} }[/tex]
⇒ E) corect
