difeffsf
春芽
共回答了17个问题采纳率:82.4% 举报
令t=x^(1/3),x=t^3,dx=3t^2dt
故原式=∫1/(t+1)*3t^2dt
=∫3(t^2-1+1)/(t+1)dt
=3∫(t-1)dt+3∫1/(t+1)dt
=3(t^2/2-t)+3ln(t+1)+C
=3(x^2/3)/2-3x^1/3+3ln|x^1/3+1|+C
1年前
追问
15
走神公
举报
麻烦你,还有这个∫√x/(√x-1)dx求不定积分,求详细步骤!
举报
difeffsf
设x^1/2=t,t^2=x,dx=2tdt 原式=∫t/(t-1)*2tdt=∫2(t^2-1+1)/(t-1)dt=2∫(t+1)dt+2∫1/(t-1)dt=t^2+2t+2ln|t-1|+C=x+2x^1/2+2ln|x^1/2-1|+C