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    Hi,

    Please can someone explain how the integral of:

    (4-2u) / (4+u^2) = 2tan^-1(u/2) - ln(4+u^2)

    I've tried by parts, looking for the log rule, but cant spot what i'm doing wrong.

    Thanks alot.
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    (Original post by Crucio404)
    Hi,

    Please can someone explain how the integral of:

    (4-2u) / (4+u^2) = 2tan^-1(u/2) - ln(4+u^2)

    I've tried by parts, looking for the log rule, but cant spot what i'm doing wrong.

    Thanks alot.
    This shows how http://www.wolframalpha.com/input/?i...1%2Bx%29%5E0.5
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    Damn, partial fractions. Bad me.
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    Isn't it a substitution I like u/4=tan(theta).
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    (4-2u) / (4+u^2)
    4/ (4+u^2) + (-2u)/ (4+u^2)

    For: (-2u)/ (4+u^2) * du use the substitution v =4+u^2 => dv/2u=du
    => (-2u)*(dv/2u)/ (v) = -1/v * dv = -lnv +C= -ln(4+u^2) +C

    The first integral is in the formula book. Hope that is clear.
 
 
 
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Updated: February 11, 2010
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