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
相关问题
