Tính  :   $I = \,\,\int\limits_{ - 1}^1 {\ln \left( {x + \sqrt {x^2 + 1} } \right)dx} $
$I = \,\,\int\limits_{ - 1}^0 {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx}  +  \,\,\int\limits_0^1 {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx} $
     $I = K + L$
   Đặt  $x =  - t\,\,\,\,\,\,\, \Rightarrow \,\,\,\,dx\,\, = \,\, - dt$
           $\begin{array}{l}
x =  - 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,t\,\, = \,\,1\\
x =  - t\,\,\,\,\,\,\, \Rightarrow \,\,\,\,t\,\, = \,\,0
\end{array}$
   Suy ra : 
  $\begin{array}{l}
K = \int\limits_{ - 1}^0 {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx = } \int\limits_0^1 {\ln \left( { - t + \sqrt {{t^2} + 1} } \right)dt} \\
\,\,\,\,\, = \,\,\,\int\limits_0^1 {\ln \frac{1}{{t + \sqrt {{t^2} + 1} }}} dt = \,\, - \int\limits_0^1 {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx =  - L}
\end{array}$
                     
   Vậy $I = K + L =  - L + L = 0$                            ĐS :  $I = 0$
ây da..tích phân đặc biệt –  nhutuyet12t7.1995 20-01-13 10:33 AM
Ad nè hoat động chăm chỉ quá –  babysexy156 14-07-12 08:14 AM
Mời bạn tham khảo ^_^ –  Tiểu Bắc 13-07-12 08:12 AM

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