区间[1/2.0]求定积分arctan2xdx

题目

区间[1/2.0]求定积分arctan2xdx

答案

应是区间 [0,1/2] 吧!
∫<0,1/2>arctan2xdx = [xarctan2x]<0,1/2>-∫<0,1/2>[2x/(1+4x^2)]dx
= π/8-(1/4)∫<0,1/2>[1/(1+4x^2)]d(1+4x^2)
= π/8-(1/4)[ln(1+4x^2)]<0,1/2>
= π/8-ln2/4 = (π-2ln2)/8.