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

Ta có \(f\left( x \right)=\int{{f}'\left( x \right)\,\text{d}x}=\int{\frac{\text{d}x}{{{x}^{2}}-1}}=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+C=\left\{ \begin{align} & \frac{1}{2}\ln \frac{x-1}{x+1}+{{C}_{1}}\,\,\,\,\,khi\,\,\,x>1 \\ & \frac{1}{2}\ln \frac{1-x}{x+1}+{{C}_{2}}\,\,\,\,\,khi\,\,\,-\,1<x<1 \\ & \frac{1}{2}\ln \frac{x-1}{x+1}+{{C}_{3}}\,\,\,\,\,khi\,\,\,x<-\,1 \\ \end{align} \right..\)

Suy ra \(f\left( -\,3 \right)+f\left( 3 \right)=0\Leftrightarrow \,\,\frac{1}{2}\ln 2+{{C}_{1}}+\frac{1}{2}\ln \frac{1}{2}+{{C}_{3}}=0\Leftrightarrow \,\,{{C}_{1}}+{{C}_{3}}=0.\) Và \(f\left( -\frac{1}{2} \right)+f\left( \frac{1}{2} \right)=2\Leftrightarrow \,\,\frac{1}{2}\ln 3+{{C}_{2}}+\frac{1}{2}\ln \frac{1}{3}+{{C}_{2}}=2\Leftrightarrow \,\,{{C}_{2}}=1.\)

Vậy \(T=f\left( -\,2 \right)+f\left( 0 \right)+f\left( 5 \right)=\frac{1}{2}\ln 3+{{C}_{3}}+{{C}_{2}}+\frac{1}{2}\ln \frac{2}{3}+{{C}_{1}}=\frac{1}{2}\ln 2+1.\)