For x > 0, \(\displaystyle e^x>1+x\). Take reciprocals to get \(\displaystyle e^{-x}<(1+x)^{-1}\). So the integrand is always negative and hence so is the integral.

For x > 0, \(\displaystyle e^x>1+x\). Take reciprocals to get \(\displaystyle e^{-x}<(1+x)^{-1}\). So the integrand is always negative and hence so is the integral.