x=2(x+1)-(x+2)
积分=∫2/(x+2)-1/(x+1)dx
=2ln|x+2|-ln|x+1|+C
x/(x^2+3x+2) = x/[(x+1)(x+2) ]
let
x/[(x+1)(x+2) ]≡A/(x+1)+B/(x+2)
=>
x≡A(x+2)+B(x+1)
x=-1, =>A=-1
x=-2, => B= 2
∫x/(x^2+3x+2) dx
=∫[-1/(x+1)+2/(x+2)] dx
=-ln|x+1| + 2ln|x+2| + C
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