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Math Help - Integral problem from Hardy book.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Integral problem from Hardy book.

    Can someone solve me this problem...
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  2. #2
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    Hint: ax^2+2bx+c=a(x^2+\frac{2b}{a}x+\frac{c}{a})=a((x+\  frac{b}{a})^2-\frac{b^2}{a^2}+\frac{c}{a})
    a((x+\frac{b}{a})^2+\frac{ac-b^2}{a^2})=a((x+\frac{b}{a})^2+(\frac{\sqrt{ac-b^2}}{a})^2)=a(\frac{\sqrt{ac-b^2}}{a})^2(\frac{(x+\frac{b}{a})^2}{(\frac{\sqrt{  ac-b^2}}{a})^2}+1)

    Now just make a proper substitution u = \frac{(x+\frac{b}{a})}{(\frac{\sqrt{ac-b^2}}{a})}
    Last edited by p0oint; June 20th 2010 at 11:42 AM. Reason: fixing LaTeX
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  3. #3
    Member mfetch22's Avatar
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    This might come in hadny considering this is an integral:

    tan^{-1}(x) = arctan(x) = \int_0^x\frac{1}{z^2+1}dz

    I'm not exactly sure how to solve this, but if you see an expression of the form \frac{1}{x^2+1}, then be on the look out for simplification by the arctangent integral indentity
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  4. #4
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    @mfetch22 you're right, check out my last post.
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