# Evaluating $\int \dfrac {1} {\sqrt{-x^{2} - 4x}}dx$

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?

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$\int \dfrac {du} {\sqrt{-u^{2} + a^{2}}}=\arcsin\dfrac{u}{a}+C$ –  user21436 Apr 7 '12 at 21:37
Ah, wow. Thanks for the catch! –  Joe Apr 7 '12 at 21:39
Thanks for the edit Brian. I was using \arcsin before someone told me to use \operatorname arcsin at one point. I always preferred \arcsin. Is there a general consensus of what to use here? –  Joe Apr 7 '12 at 21:40

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