Evaluating $ \int \frac1{\sqrt{-x^{2} - 4x}}dx$

Question

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?

Answer 1

$\displaystyle \int \dfrac{du}{\sqrt{a^2 - u^2}} = \arcsin \dfrac{u}{a} + C$

Answer 2

\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 $\int\frac{1}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}(\frac{x}{a})+c.$