Question

integral dx over square root of quantity of 2x-x^2

Answer 1

Focus on the quantity within the square root
You can rewrite 2x -x^2 as 1- (x^2 -2x +1)
this gives you
\[\int\limits_{}^{}dx / (\sqrt{1-(x-1)^{2}}\]
now put x-1=sin t
dx=cost dt
the above expression converts to costdt/[cost]
=\[\int\limits_{}^{}\]dt
=t
=\[\sin^{-1} (x-1)\]

Answer 2

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