∫（x-2）/（x^2+2x+3）^2dx
=(1/2)∫d(x^2+2x+3)/(x^2+2x+3)^2-∫3/(x^2+2x+3)^2dx
=-(1/2)(1/(x^2+2x+3))-∫3/(x^2+2x+3)^2dx
x^2+2x+3
=(x+1)^2+2
let
x+1=√2tana
dx=√2(seca)^2da
∫1/(x^2+2x+3)^2dx
=∫(1/[4(seca)^4])√2(seca)^2da
=(√2/4)∫(cosa)^2da
=(√2/8)∫(cos2a+1)da
=(√2/8)[sin2a/2+a]+C"
=(√2/8)[√2(x+1)/(x^2+2x+3)+arctan{(x+1)/2}]+C"
=-(1/2)(1/(x^2+2x+3))-(3√2/8)[√2(x+1)/(x^2+2x+3)+arctan{(x+1)/2}]+C
∫（x-2）/（x^2+2x+3）^2dx
=(1/2)∫d(x^2+2x+3)/(x^2+2x+3)^2-∫3/(x^2+2x+3)^2dx
=-(1/2)(1/(x^2+2x+3))-∫3/(x^2+2x+3)^2dx
x^2+2x+3
=(x+1)^2+2
let
x+1=√2tana
dx=√2(seca)^2da
∫1/(x^2+2x+3)^2dx
=∫(1/[4(seca)^4])√2(seca)^2da
=(√2/4)∫(cosa)^2da
=(√2/8)∫(cos2a+1)da
=(√2/8)[sin2a/2+a]+C"
=(√2/8)[√2(x+1)/(x^2+2x+3)+arctan{(x+1)/2}]+C"
∫（x-2）/（x^2+2x+3）^2dx
=-(1/2)(1/(x^2+2x+3))-∫3/(x^2+2x+3)^2dx
=-(1/2)(1/(x^2+2x+3))-(3√2/8)[√2(x+1)/(x^2+2x+3)+arctan{(x+1)/2}]+C