Biết \(\int\limits_1^4 {\sqrt {\frac{1}{{4x}} + \frac{{\sqrt x  + {e^x}}}{{\sqrt x {e^{2x}}}}} {\rm{d}}x}  = a + {e^b} - {e^c}\) vớ

* Hướng dẫn giải

Ta có \(\int\limits_1^4 {\sqrt {\frac{1}{{4x}} + \frac{{\sqrt x  + {e^x}}}{{\sqrt x {e^{2x}}}}} {\rm{d}}x}  = \int\limits_1^4 {\sqrt {\frac{{{e^{2x}} + 4x + 4{e^x}\sqrt x }}{{4x{e^{2x}}}}} {\rm{d}}x}  = \int\limits_1^4 {\sqrt {\frac{{{{\left( {{e^x} + 2\sqrt x } \right)}^2}}}{{{{\left( {2{e^x}\sqrt x } \right)}^2}}}} } {\rm{d}}x\)

\( = \int\limits_1^4 {\frac{{{e^x} + 2\sqrt x }}{{2{e^x}\sqrt x }}{\rm{d}}x}  = \int\limits_1^4 {\left( {\frac{1}{{2\sqrt x }} + \frac{1}{{{e^x}}}} \right){\rm{d}}x}  = \left( {\sqrt x  - \frac{1}{{{e^x}}}} \right)\left| \begin{array}{l}
^4\\
_1
\end{array} \right. = 1 - \frac{1}{{{e^4}}} + \frac{1}{e} = 1 + {e^{ - 1}} - {e^{ - 4}}\)

\( \to \left\{ \begin{array}{l}
a = 1\\
b =  - 1\\
c =  - 4
\end{array} \right. \to P = a + b + c =  - 4.\)