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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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5ln(1+25s^2)?
no thats not it .. the derivative of that is 5arctan(25s)... i need the anti deriv. of 5/ (1+25s^2)
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)′

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correct me if i am wrong an antideriv is going from the derivative back to the function it's self?
thanks scot.. haha i knew that .. yes thats right
zbay, yes, its also called an integral.
-5ln(u)? ,,,
Anti-derivative of that?
ya
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.
wow..... ya i wasnt even close to that answer... but ya i get it now ... thanks

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