Question

integral of 2x/(1-x^2)

Answer 1

u=1-x^2
du=-2xdx

Answer 2

yes i know, but i get -ln(1-x^2) instead of -ln(x^2-1)

Answer 3

let me try it, one sec...

Answer 4

yes, it is
-ln(1-x^2)+C
let me confer with wolfram...

Answer 5

they are the same for restricted values of x
it is because of the value restriction that that\[\ln|1-x^2|=\ln(x^2-1)\]this allows x to take on a larger range, or something... lalay can explain it better probably

Answer 6

well, I guess she left...
Anyway, they are the same because of the absolute value, and the fact that the integral is indefinite. It is the fact that the integral is indefinite that allows us to imagine the limits of evaluation to be positive, in which case ln(x^2-1) has a larger range because in ln(u), u cannot be negative. They are really the same though... I don't know why people like to do that sometimes, it seems to just be a way to drop the absolute value bars.

Answer 7

need help http://openstudy.com/#/updates/4eee0895e4b0367162f5bcb9

Answer 8

You can see that although wolfram has -ln(x^2-1) as the answer, if you click "show steps" you can see that that is just a special case.
http://www.wolframalpha.com/input/?i=integral+2x%2F%281-x%5E2%29

Answer 9

http://openstudy.com/#/updates/4eee0895e4b0367162f5bcb9

Answer 10

pls urgent