Tính tích phân : 
   $1)\,\,\,I = \int\limits_0^{\ln 3} \frac{dx}{\sqrt {e^x + 1} }        2)J = \int\limits_0^2 xe^{ - \frac{x}{2}dx} $
$1)$   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 $u = x\,\,\,\, \Rightarrow \,\,\,du = dx$
     $dv = {e^{ - \frac{x}{2}}}dx\,\,\,\, \Rightarrow \,\,v =  - 2{e^{ - \frac{x}{2}}}$
 $\begin{array}{l}
J =  - \left. {2x{e^{ - \frac{x}{2}}}} \right|_0^2 + 2\int\limits_0^2 {{e^{ - \frac{x}{2}}dx}} \\
\left. {\,\,\,\, = \, - 4{e^{ - 1}} - 4{e^{ - \frac{x}{2}}}} \right|_0^2 = \,\, - 4{e^{ - 1}} - 4\left( {{e^{ - 1}} - 1} \right) = 4 - \frac{8}{e}
\end{array}$
Vậy $J = 4 - \frac{8}{e}$
Bài này cũng đơn giản mà bạn –  babysexy156 14-07-12 08:13 AM
Thanks bạn nhiều nhé –  langtu_sitinh_dangyeuxxx 13-07-12 08:35 AM
Mời bạn xem tham khảo nhé –  Tiểu Bắc 13-07-12 08:08 AM

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