Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash...

Trên khoảng $\left( 1;+\infty \right)$ ta có $\int\limits_{{}}^{{}}{{f}'\left( x \right)dx}=\int\limits_{{}}^{{}}{\dfrac{1}{x-1}dx}=\ln \left( x-1 \right)+{{C}_{1}}\Rightarrow f\left( x \right)=\ln \left( x-1 \right)+{{C}_{1}}$.
Mà $f\left( 2 \right)=2019\Rightarrow {{C}_{1}}=2019$
Trên khoảng $\left( -\infty ;1 \right)$ ta có $\int\limits_{{}}^{{}}{{f}'\left( x \right)dx}=\int\limits_{{}}^{{}}{\dfrac{1}{x-1}dx}=\ln \left( 1-x \right)+{{C}_{2}}\Rightarrow f\left( x \right)=\ln \left( 1-x \right)+{{C}_{2}}$
Mà $f\left( 0 \right)=2018\Rightarrow {{C}_{2}}=2018$.
Vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x-1 \right)+2019 khi x>1 \\
& \ln \left( 1-x \right)+2018 khi x<1 \\
\end{aligned} \right. $. Suy ra $ f\left( 3 \right)-f\left( -1 \right)=1$
Cách khác:
Trên khoảng $\left( 1;+\infty \right)$ ta có
$\int\limits_{2}^{3}{{f}'\left( x \right)dx}=\int\limits_{2}^{3}{\dfrac{1}{x-1}dx}\Leftrightarrow f\left( x \right)\left| \begin{aligned}
& ^{3} \\
& _{2} \\
\end{aligned} \right.=\ln \left| x-1 \right|\left| \begin{aligned}
& ^{3} \\
& _{2} \\
\end{aligned} \right.\Leftrightarrow f\left( 3 \right)-f\left( 2 \right)=\ln 2 $ $ \left( 1 \right)$
Trên khoảng $\left( -\infty ;1 \right)$ ta có
$\int\limits_{-1}^{0}{{f}'\left( x \right)dx}=\int\limits_{-1}^{0}{\dfrac{1}{x-1}dx}\Leftrightarrow f\left( x \right)\left| \begin{aligned}
& ^{0} \\
& _{-1} \\
\end{aligned} \right.=\ln \left| x-1 \right|\left| \begin{aligned}
& ^{0} \\
& _{-1} \\
\end{aligned} \right.\Leftrightarrow f\left( 0 \right)-f\left( -1 \right)=-\ln 2 $ $ \left( 2 \right)$
Lấy $\left( 1 \right)+\left( 2 \right)$ vế theo vế ta được $f\left( 3 \right)-f\left( 2 \right)+f\left( 0 \right)-f\left( -1 \right)=0$
$\Rightarrow f\left( 3 \right)-f\left( -1 \right)=f\left( 2 \right)-f\left( 0 \right)=1$.