Question

求∫1/(3√x+1)dx的不定积分，式中符号为x开三次方

Answer 1

log表示ln

Answer 2

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

追问

麻烦你，还有这个∫√x/(√x-1)dx求不定积分，求详细步骤！

追答

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