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.
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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
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
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
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.
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