区间[1/2.0]求定积分arctan2xdx

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区间[1/2.0]求定积分arctan2xdx
区间[1/2.0]求定积分arctan2xdx

区间[1/2.0]求定积分arctan2xdx
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.

分部积分:∫arctan2xdx=xarctan2x-∫[2x/(1+4x²)]dx=xarctanx-∫d(x²)/(1+4x²)=
xarctan2x-0.25ln(1+4x²)+C,代入上下限得结果为π/8-0.25ln2