I=e∫21+xlnx−lnx−ln2x(1+xlnx)2dx
=e∫211+xlnxdx−e∫2lnx(lnx+1)(1+xlnx)dx
Chú ý: Nếu y=11+xlnx thì y′=−lnx+1(1+xlnx)2
* Ta tính A=−e∫2lnx(lnx+1)(1+xlnx)dx
Đặt u=lnx⇒du=dxx
dv=−lnx+1(1+xlnx)2dx⇐v=11+xlnx
⇒A=lnx1+xlnx{e2−e∫21x(1+xlnx)dx
⇒I=e∫211+xlnxdx+11+e−ln21+2ln2−e∫21x(1+xlnx)dx
⇒I=11+e−ln21+2ln2+e∫2x−1x(1+xlnx)dx
* Ta tính B=e∫2x−1x(1+xlnx)dx
B=e∫2x+xlnx−1−xlnxx(1+xlnx)dx=e∫21+lnx1+xlnxdx−e∫21xdx
Đặt t=1+xlnx⇒dt=(lnx+1)dx
x=e⇒t=1+e
x=2⇒t=1+2ln2
⇒B=1+e∫1+2ln2dtt−e∫21xdx=lnt{1+e1+2ln2−lnx{e2
⇒B=ln(1+e)−ln(1+2ln2)−1−ln2
Vậy I=11+e−ln21+2ln2+ln(1+e)−ln(1+2ln2)−1−ln2.