简单计算一下即可，答案如图所示
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第1个回答 2020-08-05
令t=[(1-x)\(1+x)]^(1\2)
则x=(1-t^2)\(1+t^2)
dx=-4tdt\(1+t^2)^2
∴∫[(1-x)\(1+x)]^(1\2)*(1\x)dx
=∫4t^2dt\(t^4-1)
=∫2dt\(t^2+1)+∫dt\(t-1)-dt\(t+1)
=2arctant+In|(t-1)\(t+1)|+C
=2arctan{[(1-x)\(1+x)]^(1\2)}+In|[1-(1-x^2)^(1\2)]\x|+C
则x=(1-t^2)\(1+t^2)
dx=-4tdt\(1+t^2)^2
∴∫[(1-x)\(1+x)]^(1\2)*(1\x)dx
=∫4t^2dt\(t^4-1)
=∫2dt\(t^2+1)+∫dt\(t-1)-dt\(t+1)
=2arctant+In|(t-1)\(t+1)|+C
=2arctan{[(1-x)\(1+x)]^(1\2)}+In|[1-(1-x^2)^(1\2)]\x|+C
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