解f(x)=(xlnx)'f(x)=lnx+1f(√x)=ln(√x)+1=(1/2)lnx+1∫f(√x)dx=∫[(1/2)lnx+1]dx=(1/2)∫lnxdx+∫dx=(1/2)xlnx-(1/2)∫xdlnx+∫dx=(1/2)xlnx-(1/2)∫dx+∫dx=(1/2)xlnx+(1/2)x+C