∫1/(X^2+2X+3)^2dx怎么求?本来原题是∫（x-2）/（x^2+2x+3）^2dx你算一下就可以到∫1/(X^2+2X+3)^2dx这步了!

Question

∫1/(X^2+2X+3)^2dx 怎么求?
本来原题是∫（x-2）/（x^2+2x+3）^2 dx 你算一下就可以到∫1/(X^2+2X+3)^2dx 这步了!

Answer 1

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