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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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Đăng bài 09-05-12 08:33 AM
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