Cho hàm số \(f\left( x \right)\) xác định trên \(R\backslash \left\{ { - 1;1} \right\}\) và thỏa mãn \(f'\left( x \right) = \frac{1}{{{x^2} - 1}}\). Biết rằng \(f\left( { - 3} \right) + f\left( 3 \right) = 0\) và \(f\left( { - \frac{1}{2}} \right) + f\left( {\frac{1}{2}} \right) = 2\). Tính \(T = f\left( { - 2} \right) + f\left( 0 \right) + f\left( 5 \right)\)

Trên khoảng \(\left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\) ta có:

 \(\begin{array}{l}f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx}  = \int\limits_{}^{} {\frac{1}{{{x^2} - 1}}dx}  = \frac{1}{2}\int\limits_{}^{} {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right)dx}  = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_1} = \frac{1}{2}\ln \frac{{x - 1}}{{x + 1}} + {C_1}\\f\left( { - 3} \right) + f\left( 3 \right) = \frac{1}{2}\ln 2 + \frac{1}{2}\ln \frac{1}{2} + {C_1} = 0 \Leftrightarrow {C_1} = 0\end{array}\)

Trên khoảng \(\left( { - 1;1} \right)\) ta có:

\(\begin{array}{l}f\left( x \right) = \int\limits_{}^{} {f'\left( x \right)dx}  = \int\limits_{}^{} {\frac{1}{{{x^2} - 1}}dx}  = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_2} = \frac{1}{2}\ln \frac{{ - x + 1}}{{x + 1}} + {C_2}\\f\left( { - \frac{1}{2}} \right) + f\left( {\frac{1}{2}} \right) = \frac{1}{2}\ln 3 + \frac{1}{2}\ln \frac{1}{3} + {C_2} = 2 \Leftrightarrow {C_2} = 2\\ \Rightarrow f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{2}\ln \frac{{x - 1}}{{x + 1}}\,\,khi\,\,x \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\\\frac{1}{2}\ln \frac{{ - x + 1}}{{x + 1}} + 2\,\,khi\,\,x \in \left( { - 1;1} \right)\end{array} \right.\\\Rightarrow T = f\left( { - 2} \right) + f\left( 0 \right) + f\left( 5 \right) = \frac{1}{2}\ln 3 + \frac{1}{2}\ln 1 + 2 + \frac{1}{2}\ln \frac{2}{3} = \ln \sqrt 2  + 2\end{array}\)