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Math Help - integrate 1/(sqrt(x^2-4))

  1. #1
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    integrate 1/(sqrt(x^2-4))

    Good afternoon,

    I managed to work my way through this problem rather quickly, but wound up with an answer that I feel is not correct (at least according to my calculator).

    I have:

     <br />
\int{\frac{1}{\sqrt{x^2-4}}dx<br />

    I evaluate to:

     <br />
\ln\mid \frac{x+\sqrt{x^2-4}}{2}\mid + C<br />

    It should be:

     <br />
\ln\mid{x+\sqrt{x^2-4}}\mid + C<br />

    I am picking up the 2 in the denominator via sec\theta = \frac{x}{2} and tan\theta = \frac{\sqrt{x^2-4}}{2}.

    Can anyone tell me what it is that I'm doing wrong here?
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  2. #2
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    Quote Originally Posted by MechEng View Post
    Good afternoon,

    I managed to work my way through this problem rather quickly, but wound up with an answer that I feel is not correct (at least according to my calculator).

    I have:

     <br />
\int{\frac{1}{\sqrt{x^2-4}}dx<br />

    I evaluate to:

     <br />
\ln\mid \frac{x+\sqrt{x^2-4}}{2}\mid + C<br />

    It should be:

     <br />
\ln\mid{x+\sqrt{x^2-4}}\mid + C<br />

    I am picking up the 2 in the denominator via sec\theta = \frac{x}{2} and tan\theta = \frac{\sqrt{x^2-4}}{2}.

    Can anyone tell me what it is that I'm doing wrong here?
    nothing wrong ...

    \displaystyle \ln\left|\frac{x+\sqrt{x^2-4}}{2}\right| + C = \ln\left|x + \sqrt{x^2-4}\right| - \ln{2} + C

    ... which only differs by a constant from the "it should be" solution.
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