求不定积分∫x^2/(x^4-x^2+1)dx

Answer 1

∫x^2dx/(x^4-x^2+1)
=(1/2)∫xdx^2/[(x^2+1)^2-3x^2]
=(1/4√3)∫(2√3x)dx^2/[(x^2+1+√3x)(x^2+1-√3x)]
=(1/4√3)[∫dx^2/(x^2+1-√3x)-∫dx^2/(x^2+1+√3x)]
=(1/4√3) [∫d(x^2-√3x+1)/(x^2+1-√3x) +∫√3dx/(x^2-√3x+1)
-∫d(x^2+√3x+1)/(x^2+1+√3x)+∫√3dx/(x^2+√3x+1)]
=(1/4√3)[ln|x^2+1-√3x|/|x^2+1+√3x| +√3∫dx/[(x-√3/2)^2+1/4]+√3∫dx/[(x+√3/2)^2+1/4]
=(1/4√3[ln|x^2+1-√3x|/|x^2+1+√3x| +2√3aratn(2x-√3)+2√3arctan(2x+√3)]+C
=(1/4√3)ln[|x^2+1-√3x|/|x^2+1+√3x|] +(1/2)arctan(2x-√3)+(1/2)arctan(2x+√3)+C