Đáp án:
\[S = f\left( 3 \right) - f\left( { - 1} \right) = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = \int {f'\left( x \right)dx} = \int {\frac{1}{{x - 1}}dx} = \ln \left| {x - 1} \right| + C\\
TH1:\,\,\,\,x < 1 \Rightarrow f\left( x \right) = \ln \left( {1 - x} \right) + {C_1}\\
f\left( 0 \right) = 2017 \Rightarrow \ln \left( {1 - 0} \right) + {C_1} = 2017 \Rightarrow {C_1} = 2017\\
\Rightarrow f\left( { - 1} \right) = \ln 2 + 2017\\
TH2:\,\,\,\,x > 1 \Rightarrow f\left( x \right) = \ln \left( {x - 1} \right) + {C_2}\\
f\left( 2 \right) = 2018 \Rightarrow \ln \left( {2 - 1} \right) + {C_2} = 2018 \Rightarrow {C_2} = 2018\\
\Rightarrow f\left( 3 \right) = \ln 2 + 2018\\
S = f\left( 3 \right) - f\left( { - 1} \right) = 1
\end{array}\)
Vậy \(S = f\left( 3 \right) - f\left( { - 1} \right) = 1\)
