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anonymous

  • 5 years ago

anti deriv. of 5/ (1+25s^2) ?

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  1. anonymous
    • 5 years ago
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    5ln(1+25s^2)?

  2. anonymous
    • 5 years ago
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    no thats not it .. the derivative of that is 5arctan(25s)... i need the anti deriv. of 5/ (1+25s^2)

  3. anonymous
    • 5 years ago
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    That looks similar to the derivative of arctan. Which is du/(1+u2)=(arctanu)′ So know that, the u would be 5s so the answer would be arctan(5s). 5/(1+(5s)2)=(arctan5s)′

  4. anonymous
    • 5 years ago
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    correct me if i am wrong an antideriv is going from the derivative back to the function it's self?

  5. anonymous
    • 5 years ago
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    thanks scot.. haha i knew that .. yes thats right

  6. anonymous
    • 5 years ago
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    zbay, yes, its also called an integral.

  7. anonymous
    • 5 years ago
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    -5ln(u)? ,,,

  8. anonymous
    • 5 years ago
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    Anti-derivative of that?

  9. anonymous
    • 5 years ago
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    ya

  10. anonymous
    • 5 years ago
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    This, I think is one of the harder integrals to see how to do it. You have to use parts. Where u = lnx and dv = 1. Then du = 1/x and v = x. So uv - integral(vdu) is xlnx - x + C is the general form, and then times the -5 so -5xlnx + 5x + C is the final answer.

  11. anonymous
    • 5 years ago
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    wow..... ya i wasnt even close to that answer... but ya i get it now ... thanks

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