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Đặt $f(x) = \left( {x + 1 - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}$
$\begin{array}{l}
t = x + \frac{1}{x}\,\, \Leftrightarrow \,\,\,\mathop {{x^2} - tx + 1}\limits_{\frac{1}{2}
\le x \le 2} \,\,\,\,\, \Leftrightarrow \,x = \left[ \begin{array}{l}
\frac{{t - \sqrt {{t^2} - 4} }}{2} \le 1\\
\frac{{t + \sqrt {{t^2} - 4} }}{2} \ge 1
\end{array} \right.\\
x = \frac{1}{2}\,\,\,\, \Rightarrow \,\,\,\,\,t = \frac{5}{2}\\
x = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,t = 2\\
x = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,t = \frac{5}{2}
\end{array}$
Suy ra: $dx = \left[ \begin{array}{l}
\frac{1}{2}\left( {1 - \frac{t}{{\sqrt {{t^2} - 4} }}}
\right)dt\,\,\,\,\,\,\,\,\,\,\\,\frac{1}{2} \le x \le 1\\
\frac{1}{2}\left( {1 + \frac{t}{{\sqrt {{t^2} - 4} }}}
\right)dt\,\,\,\,\,\,\,\,\,1 \le x \le 2
\end{array} \right.$
Từ đó: $I = \int\limits_{\frac{1}{2}}^2 {f\left( x \right)dx} = \int\limits_{\frac{1}{2}}^1 {f\left(
x \right)dx} + \int\limits_1^2 {f\left( x \right)dx} $
$\begin{array}{l}
I = \frac{1}{2}\int\limits_{\frac{5}{2}}^2 {\left( {\frac{{t - \sqrt {{t^2} - 4} }}{2} + 1 - \frac{{t +
\sqrt {{t^2} - 4} }}{2}} \right)} {e^t}.\left( {\frac{{\sqrt {{t^2} - 4} - t}}{{\sqrt {{t^2} - 4} }}}
\right)dt - \\
\,\,\, - \frac{1}{2}\int\limits_{\frac{5}{2}}^2 {\left( {\frac{{t + \sqrt {{t^2} - 4} }}{2} + 1 -
\frac{{t - \sqrt {{t^2} - 4} }}{2}} \right)} {e^t}.\left( {\frac{{\sqrt {{t^2} - 4} + t}}{{\sqrt
{{t^2} - 4} }}} \right)dt\\
\,\,\,\, = - \frac{1}{2}\int\limits_2^{\frac{5}{2}} {\left( {1 - \sqrt {{t^2} - 4} } \right)} \left( {1 -
\frac{t}{{\sqrt {{t^2} - 4} }}} \right){e^t}dt + \\
\,\,\,\,\, + \,\frac{1}{2}\int\limits_2^{\frac{5}{2}} {\left( {1 + \sqrt {{t^2} - 4} } \right)} \left( {1
+ \frac{t}{{\sqrt {{t^2} - 4} }}} \right){e^t}dt
\end{array}$
$I = \int\limits_2^{\frac{5}{2}} {\left( {\sqrt {{t^2} - 4} + \frac{t}{{\sqrt {{t^2} - 4} }}}
\right){e^t}dt} = \int\limits_2^{\frac{5}{2}} {\sqrt {{t^2} - 4} {e^t}dt + }
\int\limits_2^{\frac{5}{2}} {\frac{{t{e^t}}}{{\sqrt {{t^2} - 4} }}dt} $
Đặt $\,\,\,u = {e^{t\,\,\,\,}}\,\,\,\, \Rightarrow du = {e^t}dt$
$\,\,dv = \frac{t}{{\sqrt {{t^2} - 4} }}\,\,\,\, \Rightarrow v = \sqrt {{t^2} - 4} $
$\int\limits_2^{\frac{5}{2}} {2\frac{t}{{\sqrt {{t^2} - 4} }}} {e^t}dt = \left.
{{e^t}\sqrt {{t^2} - 4} } \right]_2^{\frac{5}{2}} - \int\limits_2^{\frac{5}{2}} {\sqrt {{t^2} - 4} }
{e^t}dt$
$1)\,\,\,I = \int\limits_0^{\ln 3} {\frac{{dx}}{{\sqrt {{e^x} + 1} }}}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2)\,\,\,J = \int\limits_0^2 {x{e^{ -
\frac{x}{2}dx}}} $
$ \Rightarrow \,\,\,\int\limits_2^{\frac{5}{2}} {\sqrt {{t^2} - 4} } {e^t}dt +
\int\limits_2^{\frac{5}{2}} {\frac{t}{{\sqrt {{t^2} - 4} }}} {e^t}dt =
\frac{3}{2}{e^{\frac{5}{2}}}$
ĐS : $I = \frac{3}{2}{e^{\frac{5}{2}}}$
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Trả lời 27-06-12 10:20 PM
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