请教定积分的换元法，一个推导，详细请见
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x^2+2x+3 = (x+1)^2 +2
let
x+1 =√2tanu
dx = √2(secu)^2 .du
-----------
∫ (x+3)/(x^2+2x+3)^2 dx
=(1/2) ∫ (2x+2)/(x^2+2x+3)^2 dx + 2∫ dx/(x^2+2x+3)^2
=-(1/2) [1/(x^2+2x+3)] +2∫ dx/(x^2+2x+3)^2
=-(1/2) [1/(x^2+2x+3)] +2∫ √2(secu)^2 .du/[ 4(secu)^4]
=-(1/2) [1/(x^2+2x+3)] +(√2/2)∫ (cosu)^2 du
=-(1/2) [1/(x^2+2x+3)] +(√2/4)∫ (1+cos2u) du
=-(1/2) [1/(x^2+2x+3)] +(√2/4)[ u +(1/2)sin2u] +C
=-(1/2) [1/(x^2+2x+3)] +(√2/4){arctan[(x+1)/√2] +√2(x+1)/(x^2+2x+3) } +C
let
x+1 =√2tanu
dx = √2(secu)^2 .du
-----------
∫ (x+3)/(x^2+2x+3)^2 dx
=(1/2) ∫ (2x+2)/(x^2+2x+3)^2 dx + 2∫ dx/(x^2+2x+3)^2
=-(1/2) [1/(x^2+2x+3)] +2∫ dx/(x^2+2x+3)^2
=-(1/2) [1/(x^2+2x+3)] +2∫ √2(secu)^2 .du/[ 4(secu)^4]
=-(1/2) [1/(x^2+2x+3)] +(√2/2)∫ (cosu)^2 du
=-(1/2) [1/(x^2+2x+3)] +(√2/4)∫ (1+cos2u) du
=-(1/2) [1/(x^2+2x+3)] +(√2/4)[ u +(1/2)sin2u] +C
=-(1/2) [1/(x^2+2x+3)] +(√2/4){arctan[(x+1)/√2] +√2(x+1)/(x^2+2x+3) } +C
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