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Math Help - Indefinite Integral U-Substitution

  1. #1
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    Indefinite Integral U-Substitution

     \int \frac {1+x^2} {4x} dx

    I'm having alot of trouble with this problem. I've tried using both 1+x^2 and 4x as U but I can't get the x's to cancel.
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  2. #2
    Master Of Puppets
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    \int \frac {1+x^2} {4x} dx= \int \frac {1} {4x} +\frac{x^2}{4x}dx= \int \frac {1} {4x} +\frac{x}{4}dx

    Any better?
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  3. #3
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    So,

     \int \frac {1} {4x} + \frac {x} {4} = \int \frac {1} {4} (x)^{-1} + \frac {1} {4} x= \frac {1} {4} ln |x| + \frac {1}{8}x^2 + c

    Yeah?
    Last edited by Matt85; June 8th 2010 at 07:28 PM. Reason: The +c will be the death me.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Matt85 View Post
    So,

     \int \frac {1} {4x} + \frac {x} {4} = \int \frac {1} {4} (x)^{-1} + \frac {1} {4} x= \frac {1} {4} ln |x| + \frac {1}{8}x^2 + c

    Yeah?
    the answer you got is correct but:


     \int \frac {1} {4x} + \frac {x} {4} = \frac{1}{4} ln(x) + \frac{1}{4} \times \frac{x^{(1+1)}}{(1+1)}+C= \frac{1}{4} ln(x) + \frac{x^2}{8} + C
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