Question

∫x²/(x²-3*x+3)²dx不定积分是多少？

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Answer 1

∫x^2/(x^2-3x+3)^2 dx
=∫dx/(x^2-3x+3) + ∫(3x-3)/(x^2-3x+3)^2 dx
=∫dx/(x^2-3x+3) + (3/2)∫(2x-3)/(x^2-3x+3)^2 dx +(3/2)∫dx/(x^2-3x+3)^2
=∫dx/(x^2-3x+3) - (3/2)[1/(x^2-3x+3)] +(3/2)∫dx/(x^2-3x+3)^2
=(2√3/3)arctan[( 2x-3)/√3] -(3/2)[1/(x^2-3x+3)]
+(2√3/3){ arctan[( 2x-3)/√3]+ √3 .(2x-3)/[4(x^2-3x+3)] }+ C
=(4√3/3)arctan[( 2x-3)/√3] +(x-3)/(x^2-3x+3)+ C
//
x^2-3x+3 = (x- 3/2)^2 + 3/4
let
x-3/2= (√3/2)tanu
dx= (√3/2)(secu)^2 du
∫dx/(x^2-3x+3)
=∫(√3/2)(secu)^2 du/ [(3/4) (secu)^2]
=(2√3/3)∫ du
=(2√3/3)u + C
=(2√3/3)arctan[( 2x-3)/√3] +C'
//
∫dx/(x^2-3x+3)^2
=∫(√3/2)(secu)^2 du/ [(9/16)(secu)^4 ]
=(8√3/9)∫ (cosu)^2 du
=(4√3/9)∫ (1+cos2u) du
=(4√3/9)[ u+(1/2)sin2u ] + C''
=(4√3/9){ arctan[( 2x-3)/√3]+ √3 .(2x-3)/[4(x^2-3x+3)] }+ C''