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ethayer Group Title

integral of (3t-2)/(t+1)

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    easiest way is probably\[u=t+1\implies t=u-1\]\[du=dt\]

    • 2 years ago
  2. TuringTest Group Title
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    long division would also work

    • 2 years ago
  3. CliffSedge Group Title
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    I tried it with long division and it was pretty easy that way.

    • 2 years ago
  4. satellite73 Group Title
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    i like the gimmick of writing \[3t-2=3t+3-5\]

    • 2 years ago
  5. ethayer Group Title
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    where do I go with long division after I get 3 remainder -5?

    • 2 years ago
  6. satellite73 Group Title
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    so you get \[\frac{3t-2}{t+1}=\frac{3t+3-5}{t+1}=1-\frac{5}{t-1}\]

    • 2 years ago
  7. satellite73 Group Title
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    oops \[\frac{3t-2}{t+1}=\frac{3t+3-5}{t+1}=3-\frac{5}{t-1}\]

    • 2 years ago
  8. satellite73 Group Title
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    then integrate each piece first one gives you \(3x\) second gives you \(-5\ln(t-1)\)

    • 2 years ago
  9. ethayer Group Title
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    ah got it thanks very much

    • 2 years ago
  10. satellite73 Group Title
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    damn another typo!! should be \[3x-5\ln(x+1)\]

    • 2 years ago
  11. RaphaelFilgueiras Group Title
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    |dw:1349062509781:dw|

    • 2 years ago
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