用换元法计算定积分∫【0到1】(x+2)/{[(x^2)+4x+1]^2 }dx

1个回答

  • ∫[0→1] (x + 2)/(x² + 4x + 1)² dx

    = ∫[0→1] (x + 2)/[(x + 2)² - 3]² dx

    令x + 2 = √3secy、dx = √3secytany dy

    x = 0 → y = arcsec(2/√3)

    x = 1 → y = arcsec(3/√3) = arcsec(√3)

    原式 = ∫ (√3secy)/(3sec²y - 3)² * [√3secytany dy]

    = ∫ (√3secy)/(9tan⁴y) * [√3secytany] dy

    = (1/3)∫ sec²y/tan³y dy

    = (1/3)∫ csc²ycoty dy

    = (1/3)∫ cscy * [cscycosy dy]

    = (- 1/3)∫ cscy d(cscy)

    = (- 1/6)csc²y

    = (- 1/6)(1 + cot²y)

    = (- 1/6)(1 + 1/tan²y)

    = (- 1/6)[1 + 1/(sec²y - 1)],上限arcsec(√3)、下限arcsec(2/√3)、分别代入

    = (- 1/6){1 + 1/[(√3)² - 1]} + (1/6){1 + 1/[(2/√3)² - 1]}

    = 5/12