计算不定积分∫x√x2+1dx
人气：128 ℃ 时间：2020-10-02 01:24:00
解答
∫x√(x^2+1)dx
=(1/2)∫√(x^2+1)dx^2
=(1/2)∫(x^2+1)^(1/2)d(x^2+1)
=(1/2)*[(x^2+1)^(1/2+1)/(1/2+1)]+C
=(1/3)(x^2+1)^(3/2)+C
=(1/3)*(x^2+1)/√(x^2+1)+C
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