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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?

  • 0
    Cool, nice to know. Th$a$nks.2012-04-07

2 Answers 2

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

  • 0
    I've removed the CW.2012-04-07
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\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 $\int\frac{1}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}(\frac{x}{a})+c.$