问题标题:
求积分∫arctanxdx
问题描述:
求积分∫arctanxdx
甘建国回答:
设t=arctanxx=tant
∫arctanxdx=∫td(tant)
=t·tant-∫tantdt
=t·tant+∫1/costd(cost)
=t·tant+ln|cost|+C
=t·tant+ln√(1-sin²t)+C
=ttant+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
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