题目

一道高数题：反常积分∫（上限正无穷,下限1）1/(x^2*(1+x))dx的值为（） A.无穷 B.0 C.ln2 D.1-ln2

提问时间：2020-11-10

答案

问题：原积分 = ∫｛x = 1 →∞｝ 1 / [ x²(1+x)] dx =
方法1：
1 / [ x²(1+x)]
= [1 - x² +x²] / [ x²(1+x)]
= [1 - x² ] / [ x²(1+x)] + x² / [ x²(1+x)]
= (1 - x) / x² + 1 / (1+x)
= [1 / x² - 1 / x + 1 / (1+x) ]
所以：原积分 = ∫｛x = 1 →∞｝ 1 / [ x²(1+x)] dx
= ∫｛x = 1 →∞｝ [1 / x² - 1 / x + 1 / (1+x) ] dx
= - 1 / x + Ln[(1+x) / x] ----------- x = 1 →∞
= 1 - Ln2 --------------- 选 D
方法2：设 x = 1 / t ｛x = 1 →∞｝ →→→→→ ｛t = 1 →0｝
原积分 = ∫｛x = 1 →∞｝ 1 / [ x²(1+x)] dx
= ∫｛t = 1 →0｝ - t / (1+t) dt
= ∫｛t = 0 →1｝ t / (1+t) dt ----------- t / (1+t) = 1 - 1 / (1 + t)
= t - Ln(1+t) t = 0 →1
= 1 - Ln2