Cho hàm số \(f(x)\) xác định trên \(R\backslash \left\{ {\frac{1}{2}} \right\},\) thỏa \(f\left( x \right) = \frac{2}{{2x - 1}},{\

* Hướng dẫn giải

Ta có \(f'\left( x \right) = \frac{2}{{2x - 1}}\)

\( \to f\left( x \right) = \int {\frac{2}{{2x - 1}}{\rm{d}}x}  = \ln \left| {2x - 1} \right| + C = \left\{ {\begin{array}{*{20}{l}}
{\ln \left( {1 - 2x} \right) + {C_1}}&{;x < \frac{1}{2}}\\
{\ln \left( {2x - 1} \right) + {C_2}}&{;x > \frac{1}{2}}
\end{array}} \right..\)

\(f\left( 0 \right) = 1 \to \ln \left( {1 - 2.0} \right) + {C_1} = 1 \to {C_1} = 1.\)

\(f\left( 1 \right) = 2 \to \ln \left( {2.1 - 1} \right) + {C_2} = 2 \to {C_2} = 2.\)

Do đó \(f\left( x \right) = \left\{ \begin{array}{l}
\ln \left( {1 - 2x} \right) + 1\,\,{\rm{khi}}\,\,x < \frac{1}{2}\\
\ln \left( {2x - 1} \right) + 2\,\,{\rm{khi}}\,\,x > \frac{1}{2}
\end{array} \right. \to \left\{ \begin{array}{l}
f\left( { - 1} \right) = \ln 3 + 1\\
f\left( 3 \right) = \ln 5 + 2
\end{array} \right.\)

\( \to f\left( { - 1} \right) + f\left( 3 \right) = 3 + \ln 5 + \ln 3 = 3 + \ln 15.\)