Originally Posted by **parallel**

tize:

the primitive of 1/1+x^2 is arctan(x) :) the primitive of 1/(1-x^2) is:

1/2 * ln[ (1+x) / (1-x) ]

but I understood the idea of what you were trying to do.

but still,i can't get the result as it should be

CaptainBlack

do you mean that the sum is just arctan(1/sqrt(3))?

i'm using an online calculator, the result should be 0.90~.

and as I said before,I can't get the result,can you please help me with my solution too?

i got that the sum is: i/sqrt(3) * 1/2 * ln[ (1+x) / (1-x) ] (x=i/sqrt(3))