Question

∫x/(1+x∧3)的不定积分

Answer 1

∫x/(1+x∧3)dx
设x/(1+x∧3)=x/[(1+x)(1-x+x^2)]=A/(1+x)+(Bx+C)/(1-x+x^2)
解得A=-1/3,B=1/3,C=1/3
∴原式=∫[(-1/3)/(1+x)+(1/3)(x+1)/(1-x+x^2)]dx
=(-1/3)ln|1+x|+(1/3)∫(1+x)/(1-x+x^2)dx
其中∫(1+x)/(1-x+x^2)dx
=(1/2)∫[(2x-1)/(1-x+x^2)+3/(1-x+x^2)]dx
=(1/2)∫1/(1-x+x^2)d(1-x+x^2)+(3/2)∫1/(1-x+x^2)dx
=(1/2)ln(1-x+x^2)+(3/2)∫1/[(x-1/2)^2+3/4]dx
=(1/2)ln(1-x+x^2)+2∫1/{[(2/√3)(x-1/2)]^2+1}dx
=(1/2)ln(1-x+x^2)+2*√3/2∫1/{[(2/√3)(x-1/2)]^2+1}d[(2/此消√3)(x-1/2)]
=(1/2)ln(1-x+x^2)+√3arctan[(2/型或√3)(x-1/2)]+C
∴∫x/(1+x∧3)dx
=(-1/3)ln|1+x|+(1/6)ln(1-x+x^2)+(√3/3)arctan[(2/√3)(x-1/卜扒伍2)]+C
=(1/6)ln[(1-x+x^2)/(1+x)^2]+(√3/3)arctan[(2x-1)/√3)]+C