integral of arccos

Question

Please help me with this definite integral. $$ \int_{-1}^1 \frac{\arccos(x)}{x^2+1}dx $$ Thanks, have a nice day!

Answer 1

In the comments, the OP has stated that the actual integral is$$I=\int\limits_{-1}^{1}dx\,\frac {\arccos x}{1+x^2}$$

Integration by parts on $u=\arccos x$ gives$$I=\arccos x\arctan x\,\Biggr\rvert_{-1}^1+\int\limits_{-1}^{1}dx\,\frac {\arctan x}{\sqrt{1-x^2}}$$Since the latter integral is odd i.e $f(x)=-f(-x)$, then it is equal to zero. Hence$$I=\arccos x\arctan x\,\Biggr\rvert_{-1}^{1}=\frac {\pi^2}4$$