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ethayer

  • 2 years ago

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

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  1. TuringTest
    • 2 years ago
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    easiest way is probably\[u=t+1\implies t=u-1\]\[du=dt\]

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

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

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

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

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

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

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

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

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

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

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