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.
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/02 15:26:57
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
![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.](/uploads/image/z/8607442-58-2.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E9%A2%98%EF%BC%9A%E5%8F%8D%E5%B8%B8%E7%A7%AF%E5%88%86%E2%88%AB%EF%BC%88%E4%B8%8A%E9%99%90%E6%AD%A3%E6%97%A0%E7%A9%B7%2C%E4%B8%8B%E9%99%901%EF%BC%891%2F%28x%5E2%2A%281%2Bx%29%29dx%E7%9A%84%E5%80%BC%E4%B8%BA%EF%BC%88%EF%BC%89+A.%E6%97%A0%E7%A9%B7+B.0+C.ln2+D.)
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.
求一道高数题答案:反常积分计算∫(上限正无穷,下限0)dx/(√ (x*(x+1)^5))的值为() A.无穷 B.0
一道高数题:反常积分∫(上限正无穷,下限0)[(b-a)x+a]dx/(2x^2+ax)的值为1,求a,b的值
求解答高数:反常积分计算∫(上限正无穷,下限0)dx/(√ (x*(x+1)^5))的值为() A.无穷 B.0 C.2
对参数p,q,讨论反常积分∫[x^p/(1+x^q)]dx的敛散性(积分下限为0,上限正无穷)
反常积分∫[上限正无穷,下限1]1 / [x√(1 - ln^2 x)]dx
求两道反常积分的计算(1)上限是正无穷 下限是1 dx/√(x(x-1))(2上限是正无穷 下限是1 dx/(x(x^2
求下限为0上限为正无穷的广义积分dx/(x^4+1)
反常积分∫x/√(1+x^2)dx 上下限是正负无穷.求敛散性?
①∫(上限为正无穷,下限为0)1/(x^2+4x+5)dx
求定积分∫ dx/x(x+1)^2 其中上限为正无穷,下限为1
定积分∫上限正无穷下限1 1/2x^2 dx