Question

求有理函数的不定积分

Answer 1

-x^2 -2

=-(x^2+x+1) +x -1

=-(x^2+x+1) +(1/2)(2x+1) - 3/2

∫ (-x^2-2)/(x^2+x+1)^2 dx

=-∫ dx/(x^2+x+1) +(1/2)∫ (2x+1)/(x^2+x+1)^2 dx -(3/2)∫ dx/(x^2+x+1)^2

=-(1/2)[1/(x^2+x+1)] -∫ dx/(x^2+x+1)  -(3/2)∫ dx/(x^2+x+1)^2

=-(1/2)[1/(x^2+x+1)] -(2√3/3)arctan[(2x+1)/√3]  

-(2√3/3) [arctan[(2x+1)/√3]+ (√3/2)(2x+1)/(x^2+x+1) ] +C

consider

x^2+x+1 = (x+1/2)^2 +3/4

let

x+1/2 =(√3/2)tanu

dx=(√3/2)(secu)^2 du

∫ dx/(x^2+x+1)

=∫ (√3/2)(secu)^2 du/ [(3/4)(secu)^2]

=(2√3/3)∫ du

=(2√3/3)u + C'

=(2√3/3)arctan[(2x+1)/√3] + C'

∫ dx/(x^2+x+1)^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+1)/√3]+ (√3/2)(2x+1)/(x^2+x+1) ] +C''