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Math Help - Harder improper integral

  1. #1
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    Harder improper integral

    The question:
    Find the \lim_{R \to \infty} \int_{-R}^{2R} \frac{x}{1 + x^2} dx

    My attempt:
    Let u = 1 + x^2

    du = 2x dx

    \frac{1}{2}\int_{-R}^{2R} \frac{du}{u}

    \frac{1}{2}_{-R}^{2R}[log(u)]

    \frac{1}{2}[log(1+(2R)^2) - log(1 + (-R)^2)]

    \frac{1}{2}[log(1+4R^2) - log(1+R^2)]

    \frac{1}{2}[log(\frac{1+4R^2}{1+R^2})]

    Limit of the function inside log is 4, so:
    \frac{1}{2}log(4)

    However, the answer is log(2). Where did I go wrong? Thanks!
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  2. #2
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    Hint: [LaTeX ERROR: Convert failed]

    Spoiler:
    \frac{1}{2}\log{4} = \log{4^{\frac{1}{2}}} = \log{2}.
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  3. #3
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    Oh dear, I can't believe I got so far and tripped up on that! Thanks!
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  4. #4
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    Quote Originally Posted by Glitch View Post
    Oh dear, I can't believe I got so far and tripped up on that! Thanks!
    Haha! It happens to us all.
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