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Math Help - Integration problems

  1. #1
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    Integration problems

    I can't figure out how to integrate the following

    2/4+x^2

    -1/(x^2+4x+14)

    I've tried them both, and on the first one, I can't get past the initial steps of u=4+x^2, du=2x.

    On the second, I assume you have to complete the square at some point, but I don't know what to do beyond that.
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  2. #2
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    Actually, in both cases, you make use of the fact that the antiderivative of \frac{1}{1+x^2} is \arctan{x}+C. Thus \int\frac{dx}{a^2+x^2} is, via the u-substitution u=\frac{x}{a}, found to be:
    \int\frac{dx}{a^2+x^2}=\int\frac{a\,du}{a^2+a^2u^2  }
    \,\,=\frac{1}{a}\int\frac{du}{1+u^2}
    \,\,=\frac{1}{a}\arctan\frac{x}{a}+C

    Both components of your integrand should be able to be transformed into multiples of \frac{1}{a^2+u^2} forms the above using appropriate u-substitutions.
    -Kevin C.
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  3. #3
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    Quote Originally Posted by quarks View Post
    [snip]
    -1/(x^2+4x+14)

    [snip]

    On the second, I assume you have to complete the square at some point, but I don't know what to do beyond that.
    -\frac{1}{x^2+4x+14} = - \frac{1}{(x + 2)^2 + 10}.

    Now make the substitution u = x + 2. Now take another look at the previous twisted reply.
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