求x5 /（x8-1)的不定积分

取u=1/x
原积分化为∫u^3/（u^8-1）du 然后把u^8-1 因式分解1/（u^4-1）（u^4+1）=
1/2[1/（u^4-1)-1/（u^4+1）]
可得两个积分 1/2∫u^3/（u^4-1）du-1/2∫u^3/（u^4+1）du
=1/8ln[（u^4-1）/（u^4+1）]
然后把x代入即可

∫x^5/(x^8-1)dx
=∫x^5/[(x^4-1)*(x^4+1)]dx
=∫x^5*1/2*[1/(x^4-1)-1/(x^4+1)]dx
=0.5∫[x^5/(x^4-1)-x^5/(x^4+1)]dx
=0.5∫{[x(x^4-1)+x]/(x^4-1)-[x(x^4+1)-x]/(x^4+1)dx
=0.5∫[x/(x^4-1)-x/(x^4+1)]dx
=0.5∫0.5x*[1/(x^2-1)-1/(x^2+1)]dx-0.5∫x/(x^4+1)dx
=0.25∫[x/(x^2-1)-x/(x^2+1)]dx-0.25∫1/(x^4+1)d(x^2)
=0.125∫1/(x^2-1)d(x^2)-0.125∫1/(x^2+1)d(x^2)-0.25arctan(x^2)
=0.125ln|x^2-1|-0.125ln(x^2+1)-0.25arctan(x^2)+c