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Xét g(t) = \sqrt t - \ln t,\,\,\,t > 0
1/\,\,\,\,\,\,\,{F^ / }(x) = {\left( {\frac{{f(x)}}{{{e^x}}}} \right)^ / } = \frac{{{f^ / }(x).{e^x} - f(x){{\left( {{e^x}} \right)}^ / }}}{{{{\left( {{e^x}} \right)}^2}}}
= \frac{{{f^ / }(x).{e^x} - f(x)\left( {{e^x}} \right)}}{{{{\left( {{e^x}} \right)}^2}}} = \frac{{{f^ / }(x) - f(x)}}{{{e^x}}}
2/\,\,\,\,\,\,\,\,f(x) = \,\, - \frac{1}{{x + 1}}\,\,\, \Rightarrow {f^ / }(x) = \frac{1}{{{{\left( {x + 1} \right)}^2}}}
{f^ / }(x) - f(x) = \frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{x + 1}} = \frac{{x + 2}}{{{{\left( {x + 1} \right)}^2}}}
3/ Ta có: \frac{{x + 2}}{{{e^x}{{\left( {x + 1} \right)}^2}}} = \frac{{{f^ / }(x) - f\left( x \right)}}{{{e^x}}}có 1 nguyên hàm của \frac{{f(x)}}{{{e^x}}}, do đó:
\int\limits_0^1 {\frac{{x + 2}}{{{e^x}{{\left( {x + 1} \right)}^2}}}dx} = \frac{{ - 1}}{{2e}} + 1 = 1 - \frac{1}{{2e}}
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