Question

求∫1/（x^2＋4x＋6）^2dx的不定积分，详细过程

Answer 1

I = ∫[1/(x^2＋4x＋6)^2]dx = ∫dx/[(x+2)^2＋2]^2
令 x+2 = √2tant， 则 dx = √2(sect)^2dt
I = ∫√2(sect)^2dt/[4(sect)^4] = (√2/4)∫(cost)^2dt
= (√2/8)∫(1+cos2t)dt = (√2/8)[t+(1/2)sin2t] + C
= (√2/8)[t+sintcost] + C
= (√2/8){arctan[(x+2)/√2]+√2(x+2)/(x^2＋4x＋6)} + C

Answer 2

原式=∫ -1/(2x+4) d[1/(x^2+4x+6)]
=-1/(2x+4)(x^2+4x+6)+(1/2)*∫ 1/(x^2+4x+6) d[1/(x+2)]
=-1/(2x+4)(x^2+4x+6)-(1/2)*∫ 1/(x^2+4x+6)(x^2+4x+4) dx
=-1/(2x+4)(x^2+4x+6)-(1/4)*∫ [1/(x^2+4x+4)-1/(x^2+4x+6)] dx
=-1/(2x+4)(x^2+4x+6)-(1/4)*∫ {1/(x+2)^2-1/[(x+2)^2+2]} dx
=-1/(2x+4)(x^2+4x+6)-(1/4)*{-1/(x+2)-(1/√2)*arctan[(x+2)/√2]}+C
=-1/(2x+4)(x^2+4x+6)+1/(4x+8)+(√2/8)*arctan[(x+2)/√2]+C
其中C是任意常数

Answer 3

x^2+4x+6 = (x+2)^2 +2
let
x+2 =√2.tanu
dx=√2.(secu)^2 du
∫ dx/(x^2+4x+6)
=∫ √2.(secu)^2 du /[2(secu)^2]
=(√2/2)∫ du
=(√2/2)u + C
=(√2/2)arctan[(x+2)/√2] + C