(1+lnx)开根号⼀(xlnx)的不定积分

2025-04-24 06:47:19
推荐回答(1个)
回答1:

设u=根号下(1+lnx),u^2=1+lnx
2udu=dx/x,代入得:
=亅2u^2/(u^2-1)*du
=2亅(1+1/(u^2-1))du
=2(u+ln|(u-1)/(u+1)|+C
=上式把u代回