令根号x=t
x=t²,dx=2tdt
x=1,t=1
x=4,t=2
所以
原式=∫(1,2) 2t/【t²(1+t)】·dt
=2∫(1,2) 1/【t(1+t)】·dt
=2∫(1,2) [1/t-1/(t+1)]dt
=2[lnt-ln(t+1)]|(1,2)
=2[ln2-ln3-ln1+ln2]
=2(2ln2-ln3)
=4ln2-2ln3