∫1/[√x+三次根号x]dx

题目

∫1/[√x+三次根号x]dx
要具体步骤 谢谢

答案

令六次根号x=t则√x=t^3三次根号x=t^2dx=6t^5dt∫1/[√x+三次根号x]dx=∫6t^5/(t^3+t^2)dt=∫6t^3/(t+1)dt=6[∫(t^3+1)/(t+1)dt-∫1/(t+1)dt]=6[∫(t^2-t+1)dt-∫1/(t+1)dt]=6[t^3/3-t^2/2+t-ln(t+1)]=2t^3-3t^2+6t-...