Answer
$$\sin ^{-1}\left(\frac{2 x-1}{3}\right)+C$$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{2+x-x^{2}}}$$ Since \begin{align*} 2+x-x^{2}&= -[x^2-x-2]\\ &=-[(x-1/2)^2-\frac{1}{4}-2]\\ &=\frac{9}{4}-(x-1/2)^2 \end{align*} Let $$x-1/2=\frac{3}{2}\sin u\ \ \ \to \ \ \ \ dx=\frac{3}{2}\cos udu$$ Then \begin{align*} \int \frac{d x}{\sqrt{2+x-x^{2}}}&=\int \frac{d x}{\sqrt{\frac{9}{4}-(x-1/2)^2}}\\ &=\int \frac{\frac{3}{2}\cos udu}{\sqrt{\frac{9}{4}-\frac{9}{4}\sin^2u}}\\ &=\int du\\ &=u+C\\ &=\sin ^{-1}\left(\frac{2 x-1}{3}\right)+C \end{align*}
