令√x=t,那么dx=2tdt
所以
原积分
=∫ [0,1] t/(1+t) *2t dt
=∫ [0,1] 2t-2 +2/(1+t) dt
=t^2 -2t +2ln|1+t| 代入上下限1和0
=1-2 +2ln2
=2ln2-1