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定积分(上下限1~0)(3x/1+x^2)dx
人气:162 ℃ 时间:2020-04-06 01:07:55
解答
d(1+x^2)=2xdx
∫[3x/(1+x^2)]dx
=3/2*∫[2x/(1+x^2)]dx
=3/2*∫d(1+x^2)/(1+x^2)
=3/2ln|1+x^2|+C
则∫[上限1,下限0][3x/(1+x^2)]dx
=3/2ln2-3/2ln1
=3/2ln2
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