题目

求根号x/(1-三次根号x)的不定积分
根号x/(1-三次根号x)的不定积分如何求?

提问时间：2021-01-02

答案

设 x^(1/6)=t,则 x=t^6,dx=6t^5dt；
∫{√x/[1- ³√x]}dx=∫[t³/(1-t²)]*6t^5 dt=6∫[t^8/(1-t²)]dt=3∫[t^8/(t-1)]dt-3∫[t^8/(t+1)]dt
=3∫[(t-1)+1]²dt/(t-1) -3∫[(t+1)-1]^8dt/(t+1)
=3∫(t-1)^7dt+3∫8(t-1)^6dt+3∫28(t-1)^5dt+3∫56(t-1)^4dt+3∫70(t-1)³dt+3∫56(t-1)²dt+3∫28(t-1)dt+3∫8dt+3∫dt/(t-1)
-3∫(t+1)^7dt+3∫8(t+1)^6dt-3∫28(t+1)^5dt+3∫56(t+1)^4dt-3∫70(t+1)³dt+3∫56(t+1)²dt-3∫28(t+1)dt+3*∫8dt-3∫dt/(t+1) +C
=(3/8)[(t-1)^8-(t+1)^8]+(24/7)[(t-1)^7+(t+1)^7]+14[(t-1)^6-(t+1)^6]+(168/5)[(t-1)^5+(t+1)^5]
+(105/2)[(t-1)^4-(t+1)^4]+56[(t-1)³+(t+1)³]+42[(t-1)²-(t+1)²]+48t+3ln[(t-1)/(t+1)] +C
再将 t=x^(1/6) 换入即得最后积分结果；