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Math Help - Tricky? Definate Integrals

  1. #1
    Senior Member polymerase's Avatar
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    Tricky? Definate Integrals

    \int^5_1 \frac{\ln x}{x^2+2x+5}\:dx

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  2. #2
    Super Member Aryth's Avatar
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    Some things are best left to the calculator:

    \int_1^5 \frac{ln(x)}{x^2 + 2x + 5}dx = 0.187 \ \text{units}^2
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  3. #3
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    Quote Originally Posted by polymerase View Post
    \int^5_1 \frac{\ln x}{x^2+2x+5}\:dx

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    Do you have reason to think that an exact solution exists?
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  4. #4
    Senior Member polymerase's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Do you have reason to think that an exact solution exists?
    This was a past test question....we are not allowed to use calculators!....so there is definalty a way to find the exact answer. I just don't know how.

    If u don't believe me, the guy said it was "0.187".....plug \frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right) and tell me what you get
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  5. #5
    Newbie Demen's Avatar
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    Using
    x = z-1

    and
    z = 2 tan Θ

    I got
    ∫ ln (2tan Θ - 1) dΘ

    but I'm not exactly sure what the next step may be.
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  6. #6
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    Quote Originally Posted by polymerase View Post
    This was a past test question....we are not allowed to use calculators!....so there is definalty a way to find the exact answer. I just don't know how.

    If u don't believe me, the guy said it was "0.187".....plug \frac{\ln5}{4}\left(\arctan3-\frac{\pi}{4}\right) and tell me what you get
    No elementary primitive exists, so it has to be handled as a definite integral from the word go. That implies special teachniques for example contour integration.

    So what we need to know is:

    What sort of mathematics is this question contexted in? The answer to this question will suggest the technique to use.
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  7. #7
    Math Engineering Student
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    Quote Originally Posted by polymerase View Post

    \int^5_1 \frac{\ln x}{x^2+2x+5}\:dx

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    Let \lambda  = \int_1^5 {\frac{{\ln x}}<br />
{{x^2  + 2x + 5}}\,dx}.

    After substitution x=\frac5u, we have \lambda  = \int_1^5 {\frac{{\ln 5 - \ln x}}<br />
{{x^2  + 2x + 5}}\,dx}.

    Hence \lambda  = \frac{1}<br />
{2}\ln 5\int_1^5 {\frac{1}<br />
{{x^2  + 2x + 5}}\,dx}. The rest follows.
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