Cho hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\backslash \left\{ -2;1 \right\}\) thỏa mãn \(f'(x)=\frac{1}{{{x}^{2}}+x-2}\);\(f(0)=\frac{1}{3}\), và\(f(-3)-f(3)=0\). Tính giá trị của biểu thức \(T=f(-4)+f(-1)-f(4).\)

Ta có \(f'\left( x \right) = \frac{1}{{{x^2} + x - 2}} \Rightarrow f\left( x \right) = \int {f'\left( x \right)\,{\rm{d}}x}  = \int {\frac{{{\rm{d}}x}}{{{x^2} + x - 2}}}  = \frac{1}{3}\ln \left| {\frac{{x - 1}}{{x + 2}}} \right| + C.\)

Khi đó \(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{3}\ln \frac{{x - 1}}{{x + 2}} + {C_1}\,\,\,\,khi\,\,\,\,x > 1\\\frac{1}{3}\ln \frac{{x - 1}}{{x + 2}} + {C_2}\,\,\,khi\,\,\,\,x <  - \,2\\\frac{1}{3}\ln \frac{{1 - x}}{{x + 2}} + {C_3}\,\,\,khi\,\,\, - \,2 < x < 1\end{array} \right.\)

Mà \(f\left( 0 \right)=\frac{1}{3}\)\( \Rightarrow \,\,{C_3} + \frac{1}{3}\ln \frac{1}{2} = \frac{1}{3} \Rightarrow {C_3} = \frac{1}{3} - \frac{1}{3}\ln \frac{1}{2}\).

Và \(f\left( -\,3 \right)-f\left( 3 \right)=0\Leftrightarrow \frac{1}{3}\ln 4+{{C}_{2}}-\frac{1}{3}\ln \frac{2}{5}-{{C}_{1}}=0\Leftrightarrow {{C}_{2}}-{{C}_{1}}=-\frac{1}{3}\ln 4+\frac{1}{3}\ln \frac{2}{5}.\)

Do đó

\(\begin{array}{l}T = \frac{1}{3}\ln \frac{5}{2} + {C_2} + \frac{1}{3}\ln 2 + {C_3} - \frac{1}{3}\ln \frac{1}{2} - {C_1}\\T = \frac{1}{3}\ln \frac{5}{2} + \frac{1}{3}\ln 2 - \frac{1}{3}\ln \frac{1}{2} - \frac{1}{3}\ln 4 + \frac{1}{3}\ln \frac{2}{5} + \frac{1}{3} - \frac{1}{3}\ln \frac{1}{2}\\T = \frac{1}{3}\ln \left( {\frac{{\frac{5}{2}.2.\frac{2}{5}}}{{\frac{1}{2}.4.\frac{1}{2}}}} \right) + \frac{1}{3}\\T = \frac{1}{3}\ln 2 + \frac{1}{3}.\end{array}\)