∫dx/[x(x^6+4)]
=∫x^2dx/[x^3*(x^6+4)]
令x^3=t--->dt=3x^2*dx--->x^2*dx=dt/3
因此原式=(1/3)∫dt/[t(t^2+4)
化有理分式1/[t(t^2+4)]為部分分式(1/4)/x-(t/4)/(t^2+4)
故前式=(1/24)∫[2dt/t-2tdt/(t^2+4)]
=(1/24)*2ln|t|-(1/24)ln(t^2+1)+C
=(1/24)ln[t^2/(t^2+4)]+C
=(1/24ln[x^6/(x^6+4)]+C.