1.f'(x)=e^x-1/(x+1),f'(0)=0,f''(x)=e^x+1/(x+1)^2>0 ,f'(x)为（-1,+∞）上的增函数,
所以x>0时,f'(x)>f'(0)=0,f(x)在（0,+∞）上是增函数,
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