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$1)$ Đặt $t = \sqrt {{e^x} + 1} \,\,\, \Rightarrow \,\,\,{t^2} = {e^x} + 1\,\,\, \Rightarrow
\,\,\,2tdt\, = {e^x}dx$
$ \Rightarrow \,\,\,dx = \frac{{2tdt}}{{{t^2} - 1}}\,\,\,;\,\,\,$$\begin{array}{l}
x = 0\,\,\, \Rightarrow \,\,t = \sqrt 2 \\
x = \ln 3\,\, \Rightarrow \,\,\,t = 2
\end{array}$
$\begin{array}{l}
I = 2\int\limits_{\sqrt 2 }^2 {\frac{{dt}}{{{t^2} - 1}}} = \left[ {\ln \left| {\frac{{t - 1}}{{t + 1}}}
\right|} \right]_{\sqrt 2 }^2 = \ln \frac{1}{3} - \ln \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}\\
I = \ln \frac{{\sqrt 2 + 1}}{{3\left( {\sqrt 2 - 1} \right)}} = \ln \frac{{3 + 2\sqrt 2 }}{3}
\end{array}$
$2)$ Đặt $u = x\,\,\,\, \Rightarrow \,\,\,du = dx$
$dv = {e^{ - \frac{x}{2}}}dx\,\,\,\, \Rightarrow \,\,v = - 2{e^{ - \frac{x}{2}}}$
$\begin{array}{l}
J = - \left. {2x{e^{ - \frac{x}{2}}}} \right]_0^2 + 2\int\limits_0^2 {{e^{ - \frac{x}{2}}}} \\
\left. {\,\,\,\, = \, - 4{e^{ - 1}} - 4{e^{ - \frac{x}{2}}}} \right]_0^2 = \,\, - 4{e^{ - 1}} - 4\left(
{{e^{ - 1}} - 1} \right) = 4 - \frac{8}{e}
\end{array}$
Vậy $J = 4 - \frac{8}{e}$
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Đăng bài 09-05-12 09:10 AM
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