<s>[INST] Evaluating $ \int \frac1{\sqrt{-x^{2} - 4x}}dx$ I am getting a sign error when evaluating:
$$ \int \dfrac {1} {\sqrt{-x^{2} - 4x}}dx$$ 
I completed the square in the denominator leaving me:
$$\int \dfrac {1} {\sqrt{-x^{2} - 4x + 4 - 4}}dx$$ 
$$\int \dfrac {1} {\sqrt{-(x^{2} + 4x - 4 + 4)}}dx$$ 
$$\int \dfrac {1} {\sqrt{-(x+2)^{2} +4}}dx$$ 
I then let $ u = x+2 , du = dx$, and $a = 2.$
$$\int \dfrac {du} {\sqrt{-u^{2} + a^{2}}}$$ 
$$\arcsin \dfrac {-(x+2)} {2} + C$$
However, the correct answer should be
$$\arcsin \dfrac {x+2} {2} + C$$
Where did I go astray?
 [/INST] \begin{align*}
\int\frac{1}{\sqrt{-x^{2}-4x}} &=\int\frac{1}{\sqrt{-x^{2}-4x+4-4}}\
                               &=\int\frac{1}{\sqrt{2^{2}-(x+2)^{2}}}\
                               &=\sin^{-1}(\frac{x+2}{2})+c.
\end{align*}
Here i use $\int\frac{1}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}(\frac{x}{a})+c.$
</s>