f'(x)=lnx，令f'(x)=lnx=0，那么x=1，∴极值点在(1,f(1))处f(x)=∫(1/2,x)lntdt=(tlnt-t)|(1/2,x)=xlnx-x+1/2*ln2+1/2∴f(1)=-1+1/2*ln2+1/2=1/2*ln2-1/2∴极值点为(1,1/2*ln2-1/2)
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