Quick Integral Check - Please Look

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  1. member910132's Avatar
    1 29-05-2012  09:27
    • Banned
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    Quick Integral Check - Please Look
    I know for a fact that
    \displaystyle \dfrac{d}{dx}(\tan^{-1}(\frac{x}{a})) = \dfrac{a}{a^2 +x^2}

    Now my FP2 textbook says
    \displaystyle \dfrac{d}{dx}(tanh^{-1}(\frac{x}{a})) = \dfrac{1}{a^2 -x^2}

    But I think
    \displaystyle \dfrac{d}{dx}(tanh^{-1}(\frac{x}{a})) = \dfrac{a}{a^2 -x^2}

    Am I right ?

    Thnx
    Last edited by member910132; 29-05-2012 at 09:46.
  2. nuodai's Avatar
    2 29-05-2012  09:29
    • PS Helper
    • TSR Legend
    Re: Quick Integral Check - Please Look
    You are correct.
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