Integrate the function: $\sqrt{1-4x-x^{2}}$

Let $I = \int \sqrt{1-4x-x^{2}} dx$.
To integrate this,we complete the square inside the square root:
$1-4x-x^{2} = 1 - (x^{2} + 4x) = 1 - (x^{2} + 4x + 4 - 4) = 1 - ((x+2)^{2} - 4) = 5 - (x+2)^{2}$.
So,$I = \int \sqrt{(\sqrt{5})^{2} - (x+2)^{2}} dx$.
Using the standard integral formula $\int \sqrt{a^{2} - t^{2}} dt = \frac{t}{2} \sqrt{a^{2} - t^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{t}{a}) + C$,where $t = x+2$ and $a = \sqrt{5}$:
$I = \frac{x+2}{2} \sqrt{5 - (x+2)^{2}} + \frac{5}{2} \sin^{-1}(\frac{x+2}{\sqrt{5}}) + C$.
Simplifying the term under the square root back to the original form:
$I = \frac{x+2}{2} \sqrt{1-4x-x^{2}} + \frac{5}{2} \sin^{-1}(\frac{x+2}{\sqrt{5}}) + C$,where $C$ is an arbitrary constant.