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Math Help - Integration Query

  1. #1
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    Integration Query

    9cii please!

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  2. #2
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    Quote Originally Posted by Kevlar View Post
    9cii please!

    \int u \, dv = uv - \int v \, du.

    Let u = \ln (4x - 3) and dv = dx.
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  3. #3
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    I've gotten upto

    Integral ln4x-3 dx = xln4x-3 - integral x X 4/4x-3

    then i get stuck :\
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  4. #4
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    \int{\ln (4x-3)\,dx}=\int{(x)'\ln (4x-3)\,dx}=x\ln (4x-3)-4\int{\frac{x}{4x-3}\,dx}.

    See my signature for further details about integration by parts.
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  5. #5
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    Quote Originally Posted by Kevlar View Post
    I've gotten upto

    Integral ln4x-3 dx = xln4x-3 - integral x X 4/4x-3

    then i get stuck :\
    Note that \frac{4x}{4x - 3} = \frac{(4x - 3) + 3}{4x-3} = 1 + \frac{1}{4x - 3}.
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  6. #6
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    \int ln(4x-3)dx

    \int ln(4x-3).dx
    . means multiply okok
    u= ln(4x-3) du/dx= 4/4x-3
    dV/dx= 1, V= x
    sooooo
    \int ln(4x-3)dx = xln(4x-3) - \int x.\frac{4}{4x-3}
    \int ln(4x-3)dx = xln(4x-3) - \int \frac{4x}{4x-3}
    second integral look familiar ? Its the question before with substitution how lovely!
    \int ln(4x-3)dx = xln(4x-3) - \frac{4x-2}{4}+ \frac{3}{4}ln(4x-3) + C
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