Cho hàm số $f(x)$ xác định trên $\mathbb{R}\backslash \left\{ -1;2...

Ta có: $f(x)=\int{{f}'(x)\text{d}x}=\int{\dfrac{\text{d}x}{{{x}^{2}}-x-2}}=\dfrac{1}{3}\int{\left( \dfrac{1}{x-2}-\dfrac{1}{x+1} \right)\text{d}x}$ $=\dfrac{1}{3}\left( \ln \left| x-2 \right|-\ln \left| x+1 \right| \right)+C$
Suy ra $f(x)=\left\{ \begin{aligned}
& \dfrac{1}{3}\left( \ln (2-x)-\ln (-x-1) \right)+{{C}_{1}} khi x<-1 \\
& \dfrac{1}{3}\left( \ln (2-x)-\ln (x+1) \right)+{{C}_{2}} khi -1<x<2 \\
& \dfrac{1}{3}\left( \ln (x-2)-\ln (x+1) \right)+{{C}_{3}} khi x>2 \\
\end{aligned} \right.$
$\Rightarrow f(-3)=\dfrac{1}{3}\left( \ln 5-\ln 2 \right)+{{C}_{1}}$, $f(3)=\dfrac{1}{3}\left( \ln 1-\ln 4 \right)+{{C}_{3}}=\dfrac{-2}{3}\ln 2+{{C}_{3}}$
$\Rightarrow f(-3)-f(3)=\dfrac{1}{3}\left( \ln 5-\ln 2 \right)+{{C}_{1}}+\dfrac{2}{3}\ln 2-{{C}_{3}}=0$
$\Leftrightarrow \dfrac{1}{3}\ln 10+{{C}_{1}}-{{C}_{3}}=0\Leftrightarrow {{C}_{1}}-{{C}_{3}}=-\dfrac{1}{3}\ln 10$
$f(0)=\dfrac{1}{3}\ln 2+{{C}_{2}}=\dfrac{1}{3}\Rightarrow {{C}_{2}}=\dfrac{1}{3}-\dfrac{1}{3}\ln 2$
$f(-4)=\dfrac{1}{3}\left( \ln 6-\ln 3 \right)+{{C}_{1}}=\dfrac{1}{3}\ln 2+{{C}_{1}}$
$f(4)=\dfrac{1}{3}\left( \ln 2-\ln 5 \right)+{{C}_{3}}$
$f(1)=\dfrac{1}{3}\left( \ln 1-\ln 2 \right)+{{C}_{2}}=-\dfrac{1}{3}\ln 2+{{C}_{2}}$
Suy ra $f(-4)+f(1)-f(4)=\dfrac{1}{3}\ln 2+{{C}_{1}}-\dfrac{1}{3}\ln 2+{{C}_{2}}-\dfrac{1}{3}\left( \ln 2-\ln 5 \right)-{{C}_{3}}$
$=-\dfrac{1}{3}\ln 10+\dfrac{1}{3}-\dfrac{1}{3}\ln 2-\dfrac{1}{3}\left( \ln 2-\ln 5 \right)=\dfrac{1}{3}-\ln 2$