Question

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.

Answer 1

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