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Thread: Integration Question (inverse trig)

  1. Mar 10th 2010, 06:42 PM #1
    jfinkbei
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    Integration Question (inverse trig)

    Question:
    \int \frac{dx}{x^2+6x+10}

    If it was -6x I believe the method is to complete the square but not sure with +6x...

    Would like some tips please
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  2. Mar 10th 2010, 06:51 PM #2
    Black
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    You would still complete the square.

    x^2+6x+10=(x+3)^2+1, so we have

    \int \frac{dx}{(x+3)^2+1}=\int \frac{du}{u^2+1} (letting u=x+3).

    Can you go from here?
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  3. Mar 10th 2010, 06:53 PM #3
    jfinkbei
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    Yah I got it, thanks a lot I have been feeling dumb a lot lately after some of my posts here...
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