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求1/(x^2-x+1)^3/2在(0,1)上的定积分
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求1/(x^2-x+1)^3/2在(0,1)上的定积分
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答案和解析
答案是4/3.用第二换元法
∫(0→1) 1/(x² - x + 1)^(3/2) dx
= ∫(0→1) 1/[(x - 1/2)² + 3/4]^(3/2) dx
令x - 1/2 = (√3/2)tanz、dx = (√3/2)sec²z dz
√[(x - 1/2)² + 3/4] = √[(3/4)tan²z + 3/4] = √[(3/4)sec²z] = (√3/2)secz
x = 0 ===> tanz = - 1/√3 ===> z = - π/6
x = 1 ===> tanz = 1/√3 ===> z = π/6
原式 = ∫(- π/6→π/6) 1/[(√3/2)secz]³ * [(√3/2)sec²z dz]
= (4/3)∫(- π/6→π/6) cosz dz
= (4/3)[sinz] |(- π/6→π/6)
= (4/3)[(1/2) - (- 1/2)]
= 4/3
∫(0→1) 1/(x² - x + 1)^(3/2) dx
= ∫(0→1) 1/[(x - 1/2)² + 3/4]^(3/2) dx
令x - 1/2 = (√3/2)tanz、dx = (√3/2)sec²z dz
√[(x - 1/2)² + 3/4] = √[(3/4)tan²z + 3/4] = √[(3/4)sec²z] = (√3/2)secz
x = 0 ===> tanz = - 1/√3 ===> z = - π/6
x = 1 ===> tanz = 1/√3 ===> z = π/6
原式 = ∫(- π/6→π/6) 1/[(√3/2)secz]³ * [(√3/2)sec²z dz]
= (4/3)∫(- π/6→π/6) cosz dz
= (4/3)[sinz] |(- π/6→π/6)
= (4/3)[(1/2) - (- 1/2)]
= 4/3
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