∫(x-2)√(x^2 4x 1) dx

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∫(x-2)√(x^2 4x 1) dx
求不定积分,
∫(x-2)√(x^2+4x+1)

∫(x-2)√(x^2+ 4x+ 1) dx
= (1/2)∫(2x+4)√(x^2+ 4x+ 1) dx - 4∫√(x^2+ 4x+ 1) dx
= (1/3)(x^2+ 4x+ 1)^(3/2) - 4∫√(x^2+ 4x+ 1) dx
consider
x^2+4x+1 = (x+2)^2 - 3
let
x+2 = √3secy
dx = √3secytany dy
∫√(x^2+ 4x+ 1) dx
=3∫(secy)^2 .tany dy
=3∫secy dsecy
=(3/2)(secy)^2 + C'
=(1/2)(x+2)^2 + C'
∫(x-2)√(x^2+ 4x+ 1) dx
= (1/3)(x^2+ 4x+ 1)^(3/2) - 4∫√(x^2+ 4x+ 1) dx
= (1/3)(x^2+ 4x+ 1)^(3/2) - 2(x+2)^2 + C
再问： ∫√(x^2+ 4x+ 1) dx
=3∫(secy)^2 .tany dy这一步有错误吧
再答： ∫(x-2)√(x^2+ 4x+ 1) dx
= (1/2)∫(2x+4)√(x^2+ 4x+ 1) dx - 4∫√(x^2+ 4x+ 1) dx
= (1/3)(x^2+ 4x+ 1)^(3/2) - 4∫√(x^2+ 4x+ 1) dx
consider
x^2+4x+1 = (x+2)^2 - 3
let
x+2 = √3secy
dx = √3secytany dy

∫√(x^2+ 4x+ 1) dx
=3∫(secy)(tany)^2 dy

=3∫(secy)[(secy)^2 -1 ]dy
= 3∫(secy)^3 dy - ln|secy + tany |
consider
∫(secy)^3 dy = ∫(secy) dtany
= secy tany - ∫(tany)^2.secy dy
2∫(secy)^3 dy =secy tany + ∫secy dy
∫(secy)^3 dy =(1/2)[ secy tany + ln|secy+tany| ]

∫√(x^2+ 4x+ 1) dx
= 3∫(secy)^3 dy - ln|secy + tany |
=(3/2)[ secy tany + ln|secy+tany| ] - ln|secy + tany |
=(1/2)[ 3secy tany + ln|secy+tany| ]
=(1/2){ (x+2) .√(x^2+4x+1) + ln|( x+2+ √(x^2+4x+1) ) /√3| ]

∫(x-2)√(x^2+ 4x+ 1) dx
= (1/3)(x^2+ 4x+ 1)^(3/2) - 4∫√(x^2+ 4x+ 1) dx
= (1/3)(x^2+4x+1)^(3/2) - 2[ (x+2).√(x^2+4x+1) + ln| [ x+2+√(x^2+4x+1) ] /√3| ] + C

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