Question

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

Answer 1

5ln(1+25s^2)?

Answer 2

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

Answer 3

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)′

Answer 4

correct me if i am wrong an antideriv is going from the derivative back to the function it's self?

Answer 5

thanks scot.. haha i knew that .. yes thats right

Answer 6

zbay, yes, its also called an integral.

Answer 7

-5ln(u)? ,,,

Answer 8

Anti-derivative of that?

Answer 9

ya

Answer 10

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.

Answer 11

wow..... ya i wasnt even close to that answer... but ya i get it now ... thanks