区间[1/2.0]求定积分arctan2xdx

分类: 作业答案 • 2022-06-19 20:43:21

问题描述：

答

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