evaluate integral( log(1+x)/(1+x^2)) here can we use log(1+x) series

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- March 16th 2012, 03:44 AMprasumcalculus
evaluate integral( log(1+x)/(1+x^2)) here can we use log(1+x) series

- March 16th 2012, 05:26 AMHallsofIvyRe: calculus
Okay, what

**is**that series? - March 17th 2012, 02:43 AMprasumRe: calculus
the series is x-x^2/2+x^3/3_x^4/4... infinite