Evaluate the integral $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$.

We have the integral $I = \int_{0}^{2} \frac{d x}{x+4-x^{2}}$.
First,complete the square in the denominator: $x+4-x^{2} = -(x^{2}-x-4) = -\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4\right) = -\left(\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right) = \left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}$.
Now,the integral becomes $I = \int_{0}^{2} \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}$.
Let $t = x-\frac{1}{2}$,then $dt = dx$. When $x=0, t=-\frac{1}{2}$ and when $x=2, t=\frac{3}{2}$.
Using the formula $\int \frac{dx}{a^{2}-x^{2}} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,we get:
$I = \left[ \frac{1}{2(\frac{\sqrt{17}}{2})} \log \left| \frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t} \right| \right]_{-\frac{1}{2}}^{\frac{3}{2}} = \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17}+2t}{\sqrt{17}-2t} \right| \right]_{-\frac{1}{2}}^{\frac{3}{2}}$.
Substituting the limits: $I = \frac{1}{\sqrt{17}} \left( \log \left| \frac{\sqrt{17}+3}{\sqrt{17}-3} \right| - \log \left| \frac{\sqrt{17}-1}{\sqrt{17}+1} \right| \right) = \frac{1}{\sqrt{17}} \log \left( \frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1} \right)$.
Simplifying the argument: $\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)} = \frac{17+\sqrt{17}+3\sqrt{17}+3}{17-\sqrt{17}-3\sqrt{17}+3} = \frac{20+4\sqrt{17}}{20-4\sqrt{17}} = \frac{5+\sqrt{17}}{5-\sqrt{17}}$.
Rationalizing: $\frac{(5+\sqrt{17})^{2}}{25-17} = \frac{25+17+10\sqrt{17}}{8} = \frac{42+10\sqrt{17}}{8} = \frac{21+5\sqrt{17}}{4}$.
Thus,$I = \frac{1}{\sqrt{17}} \log \left( \frac{21+5\sqrt{17}}{4} \right)$.