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设t=x+2,
原式=∫(t-2)dt/(t3-t)
=∫dt/(t2-1)-2∫dt/(t3-t)
=(1/2)ln[(t-1)/(t+1)]-∫dt2/[t2(t2-1)]
=(1/2)ln[(x+1)/(x+3)]-∫d(t2-1/2)/[(t2-1/2+1/2)(t2-1/2-1/2)]
=(1/2)ln[(x+1)/(x+3)]-ln[(t2-1/2-1/2)/(t2-1/2+1/2)]+C
=[ln(x+1)]/2-[ln(x+3)]/2-ln(x2+4x+3)+2ln(x+2)+C