Biết \(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\frac{1}{{\sqrt {{e^{2x}} + 1}  - {e^x}}}{\rm{d}}x}  = 1 + \frac{1}{2}\ln \frac{b}{a} +

* Hướng dẫn giải

Ta có \(I = \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\frac{1}{{\sqrt {{e^{2x}} + 1}  - {e^x}}}{\rm{d}}x}  = \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\left( {\sqrt {{e^{2x}} + 1}  + {e^x}} \right){\rm{d}}x}  = \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\sqrt {{e^{2x}} + 1} {\rm{d}}x}  + \int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {{e^x}{\rm{d}}x} .\)

\(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {{e^x}{\rm{d}}x}  = {e^x}\left| \begin{array}{l}
^{\ln \sqrt 8 }\\
_{\ln \sqrt 3 }
\end{array} \right. = 2\sqrt 2  - \sqrt 3 .\)

\(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\sqrt {{e^{2x}} + 1} {\rm{d}}x} .\) Đặt \(t = \sqrt {{e^{2x}} + 1}  \Leftrightarrow {t^2} = {e^{2x}} + 1\) suy ra \(2t{\rm{d}}t = 2{e^{2x}}{\rm{d}}x \Leftrightarrow {\rm{d}}x = \frac{{t{\rm{d}}t}}{{{e^{2x}}}} = \frac{{t{\rm{d}}t}}{{{t^2} - 1}}.\)

Đổi cận \(\left\{ \begin{array}{l}
x = \ln \sqrt 3  \to t = 2\\
x = \ln \sqrt 8  \to t = 3
\end{array} \right..\)

Khi đó \(\int\limits_{\ln \sqrt 3 }^{\ln \sqrt 8 } {\sqrt {{e^{2x}} + 1} {\rm{d}}x}  = \int\limits_2^3 {\frac{{{t^2}{\rm{d}}t}}{{{t^2} - 1}}} dt = \int\limits_2^3 {\left( {1 + \frac{1}{{{t^2} - 1}}} \right){\rm{d}}t}  = \left( {t + \frac{1}{2}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right)\left| \begin{array}{l}
^3\\
_2
\end{array} \right. = 1 + \frac{1}{2}\ln \frac{3}{2}.\)

Vậy \(I = 1 + \frac{1}{2}\ln \frac{3}{2} + 2\sqrt 2  - \sqrt 3  \to \left\{ \begin{array}{l}
a = 2\\
b = 3
\end{array} \right. \to P = a + b = 5.\)