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anonymous

  • 5 years ago

does anyone know how to find the integral of ln(2x+1)

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  1. anonymous
    • 5 years ago
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    Make a substitution of u=(2x+1), find du and use integration by parts.

  2. anonymous
    • 5 years ago
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    so i end up wiht an integral of 1/2ln(u).du, but then what?

  3. anonymous
    • 5 years ago
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    u=2x+1 then du=2dx then dx = du/2, so, letting your integral be I,\[I=\frac{1}{2}\int\limits_{}{}\ln u du\]

  4. anonymous
    • 5 years ago
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    \[I=\frac{1}{2}\int\limits_{}{}\ln w dw\]I've just made a switch on the dummy variable since I want to use convention to integrate with IBP, and this convention uses u.

  5. anonymous
    • 5 years ago
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    yeah, i used y for the same reason

  6. anonymous
    • 5 years ago
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    \[ u=\ln w \rightarrow du = \frac{dw}{w}\]and\[dv=dw \rightarrow v=w\]Then\[2I=\ln w .w-\int\limits_{}{}w \frac{dw}{w}=w \ln w - w + c\]Since \[w=2x+1\]\[2I=(2x+1)\ln (2x+1) -(2x+1)+c\]Divide by 2 to get I.

  7. anonymous
    • 5 years ago
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    yeah, thats what i got, teh answer in the back of my text book gives \[1/2(2x+1)\ln (2x+1)-x+C\] but i guess they just expanded the last (2x+1)/2 and included the 1/2 with C

  8. anonymous
    • 5 years ago
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    most likley

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