The integral is #int_0^1ln(x)\ dx#.
Use integration by parts #int\ u\ dv=uv-int\ v\ du# where #u=ln(x),du=1/x,dv=dx,v=x#.
#=[xln(x)]_0^1-int_0^1dx#
#=[xln(x)-x]_0^1#
#=(1*ln(1)-1)-(0*ln(0)-0)#
#=-1-(0*ln(0)-0)#
The problem is that #ln(0)# is undefined. Use limits instead:
#=-1-lim_(x->0)(xln(x)-x)#
#=-1-lim_(x->0)(xln(x))#
#=-1-lim_(x->0)(ln(x)/(1/x))#
Use l'Hopital's rule:
#=-1-lim_(x->0)((1/x)/(-1/x^2))#
#=-1-lim_(x->0)((1/x)/(-1/x^2))#
#=-1-lim_(x->0)(-x)#
#=-1#
Here is another way to do the integral. Observe that the area is equivalent to the area below #y=e^x# (the inverse function) from #x=-oo# to #0# but opposite in sign.
#-int_(-oo)^0e^x\ dx#
#=-[e^x]_(-oo)^0#
#=-1# since #lim_(x->-oo)e^x=0#
