1/2掉了，改下
可以采纳么
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第1个回答 2015-01-06
令t=tanx x=arctant dx=dt/(1+t^2)
原式=∫arctant*tdt
=1/2*∫arctant d(t^2)
=1/2*[arctant*t^2-∫t^2d(arctant)]
=t^2/2*arctant-1/2*∫t^2/(1+t^2)dt
=t^2/2*arctant-1/2*∫[1-1/(1+t^2)]dt
=t^2/2*arctant-1/2*(t-arctant)+C
=(t^2+1)/2*arctant-t/2+C
=(tan^2x+1)x/2-(tanx)/2+C
原式=∫arctant*tdt
=1/2*∫arctant d(t^2)
=1/2*[arctant*t^2-∫t^2d(arctant)]
=t^2/2*arctant-1/2*∫t^2/(1+t^2)dt
=t^2/2*arctant-1/2*∫[1-1/(1+t^2)]dt
=t^2/2*arctant-1/2*(t-arctant)+C
=(t^2+1)/2*arctant-t/2+C
=(tan^2x+1)x/2-(tanx)/2+C
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