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experimentX

  • 2 years ago

Integrate: \[ \int_0^\infty {\ln x \over 1+x^2}dx\]

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  1. hba
    • 2 years ago
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    I cannot see latex :(

  2. experimentX
    • 2 years ago
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    |dw:1355769569794:dw|

  3. experimentX
    • 2 years ago
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    it's pretty easy after you know it's value. If possible try without knowing the final results.

  4. Zarkon
    • 2 years ago
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    look at \[\int_{1}^{\infty}\frac{\log(x)}{1+x^2}dx\] make the substitution \(x=\frac{1}{u}\)

  5. hba
    • 2 years ago
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    So 0 is the final result.

  6. experimentX
    • 2 years ago
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    yep

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