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eexiam

  • 3 years ago

integrate by parts: ∫arctan4tdt

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  1. TuringTest
    • 3 years ago
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    \[u=\tan^{-1}(4t)\]\[dv=tdt\]

  2. tjones89
    • 3 years ago
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    i think you saw an extra t in there dv should just be dt

  3. TuringTest
    • 3 years ago
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    yes I did :)

  4. TuringTest
    • 3 years ago
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    \[u=\tan^{-1}(4t)\]\[dv=dt\]I'm tired now, g'night!

  5. tjones89
    • 3 years ago
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    \[du = \frac{ 4 }{ 1+t^2 }dt\] \[v = t\]

  6. tjones89
    • 3 years ago
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    \[I = t*\arctan(4t) - 4\int\limits_{}^{}\frac{ t }{ 1+t^2 }dt\] u-sub from here should do it

  7. Fellowroot
    • 3 years ago
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    |dw:1350628982821:dw|

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