∫﹙1,e³﹚1/[x√﹙1＋lnx﹚]dx
=∫﹙1,e³﹚1/√﹙1＋lnx﹚]dlnx
=∫﹙1,e³﹚1/√﹙1＋lnx﹚]d(lnx+1)
=2√﹙1＋lnx﹚﹙1,e³﹚
=4-2
=2