利用分部积分法求{ln[x+(x*x-1)]}的不定积分.
人气：220 ℃ 时间：2020-06-10 13:10:06
解答
∫[log(x + (x^2 - 1)^(1/2))]dx=x*log(x + (x^2 - 1)^(1/2)) -∫x*d[log(x + (x^2 - 1)^(1/2))]=x*log(x + (x^2 - 1)^(1/2))- ∫x/( (x^2 - 1)^(1/2))*dx=x*log(x + (x^2 - 1)^(1/2)) - (x^2 - 1)^(1/2)
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