∫1/[x(x^6+4)]dx

Question

∫1/ [x(x^6+4)] dx

Answer 1

因为d(x^3)=3x^2dx
所以dx=d(x^3)/3(x^2)
原式=∫1/[(x^6+4)*x*3x^2] dx
=∫1/[(x^6+4)*3x^3] d(x^3)
令x^3=t
原式=(1/3)∫1/[(t^2+4)t] d(t)
=(1/3)∫1/[(t^2+4)t*2t] d(t^2)
=(1/6)∫1/[(t^2+4)t^2] d(t^2)
=(1/24)∫(1/t^2)-[1/(t^2+4)] d(t^2)
=(1/24)∫(1/t^2)d(t^2)-(1/24)∫1/(t^2+4)d(t^2)
=(lnt^2)/24-[ln(t^2+4)]/24
所以原式=(lnx^6)/24-[ln(x^6+4)]/24
=(lnx)/4-[ln(x^6+4)]/24