$$\int \frac{(x+1)dx}{\sqrt{2x-x^2}}$$
completing the square:
$$1-(x^2-2x+1)$$ $$1-(x+1)^2$$ $$a^2-U^2$$
we have: $$U = (x+1) $$ $$a = 1$$
we have:
$$\int \frac{UdU}{\sqrt{a^2-U^2}}$$
let: $ U= a\sin\theta$ and $dU = a\cos\theta d\theta$ hence:
$$\int \frac{a\sin\theta a\cos\theta}{a\cos\theta}$$ $$ = -a\cos\theta$$ $$ = -\sqrt{2x-x^2}$$
Which is wrong, what did I miss?
It's $1-(x-1)^2$ rather than $1-(x+1)^2$