∫(√x-1)/(³√x+1)dx
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∫(√x-1)/(³√x+1)dx
∫(√x-1)/(³√x+1)dx
∫(√x-1)/(³√x+1)dx
令六次根号x=t
x=t^6
dx=6t^5dt
代入得
∫(√x-1)/(³√x+1)dx
=∫(t^3-1)/(t^2+1)*6t^5dt
=6∫(t^8-t^5)/(t^2+1)dt
=6∫[t^6-t^4-t^3+t^2+t-1-t/(t^2+1)-1/(t^2+1)]dt
自己积分吧