求积分∫arctanxdx
1个回答

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