求积分∫arctanxdx

Question

求积分∫arctanxdx

Answer 1

设t=arctanx x=tant
∫arctanxdx = ∫ td(tant)
=t·tant - ∫ tantdt
=t·tant + ∫ 1/cost d(cost)
=t·tant + ln|cost| + C
=t·tant + ln√(1-sin²t) + C
=t tant + ln√[1-(1-cos2t)/2] + C
=t·tant + ln√[1/2(1+cos2t)] +C
=t· tant + ln√1/2[1+(1-tan²t)/(1+tan²t)] + C
带入t=arctanx 有
∫arctanxdx =xarctanx + ln√1/(1+x²) + C