∫lnx(1+lnx)2dx
Put lnx=t⟹x=et
⟹dx=etdt
=∫t.et(1+t)2dt
=∫et×1(1+t)dt−∫et.1(1+t)2dt
=1(1+t)et+∫1(1+t)2etdt−∫etdt(1+t)2
=et1+t+c
=x1+lnx+c