Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f'\left( x \right) = \dfrac{4}{{{x^2} - 4}} = \dfrac{4}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{\left( {x + 2} \right) - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{x - 2}} - \dfrac{1}{{x + 2}}\\
\Rightarrow f\left( x \right) = \int {f'\left( x \right)dx} = \int {\left( {\dfrac{1}{{x - 2}} - \dfrac{1}{{x + 2}}} \right)dx} \\
\Rightarrow f\left( x \right) = \ln \left| {x - 2} \right| - \ln \left| {x + 2} \right| + C\\
TH1:\,\,\,\,x < - 2 \Rightarrow \left\{ \begin{array}{l}
x - 2 < 0\\
x + 2 < 0
\end{array} \right.\\
\Rightarrow f\left( x \right) = \ln \left( {2 - x} \right) - \ln \left( { - x - 2} \right) + {C_1}\\
f\left( { - 3} \right) = 0 \Leftrightarrow \ln \left( {2 - \left( { - 3} \right)} \right) - \ln \left( { - \left( { - 3} \right) - 2} \right) + {C_1} = 0\\
\Leftrightarrow \ln 5 - \ln 1 + {C_1} = 0\\
\Rightarrow {C_1} = - \ln 5\\
TH2:\,\,\,\, - 2 \le x \le 2 \Rightarrow \left\{ \begin{array}{l}
x - 2 \le 0\\
x + 2 > 0
\end{array} \right.\\
\Rightarrow f\left( x \right) = \ln \left( {2 - x} \right) - \ln \left( {x + 2} \right) + {C_2}\\
f\left( 0 \right) = 1 \Leftrightarrow \ln \left( {2 - 0} \right) - \ln \left( {0 + 2} \right) + {C_2} = 1\\
\Leftrightarrow \ln 2 - ln2 + {C_2} = 1\\
\Leftrightarrow {C_2} = 1\\
TH3:\,\,\,\,x > 2 \Rightarrow \left\{ \begin{array}{l}
x - 2 > 0\\
x + 2 > 0
\end{array} \right.\\
\Rightarrow f\left( x \right) = \ln \left( {x - 2} \right) - \ln \left( {x + 2} \right) + {C_3}\\
f\left( 3 \right) = 2 \Leftrightarrow \ln \left( {3 - 2} \right) - \ln \left( {3 + 2} \right) + {C_3} = 2\\
\Leftrightarrow \ln 1 - \ln 5 + {C_3} = 2\\
\Rightarrow {C_3} = 2 + \ln 5\\
f\left( { - 4} \right) + f\left( { - 1} \right) + f\left( 4 \right)\\
= \left[ {\ln \left( {2 - \left( { - 4} \right)} \right) - \ln \left( { - 2 - \left( { - 4} \right)} \right) + {C_1}} \right] + \left[ {\ln \left( {2 - \left( { - 1} \right)} \right) - \ln \left( { - 1 + 2} \right) + {C_2}} \right] + \left[ {\ln \left( {4 - 2} \right) - \ln \left( {4 + 2} \right) + {C_3}} \right]\\
= \left( {\ln 6 - \ln 2 - \ln 5} \right) + \left( {\ln 3 - \ln 1 + 1} \right) + \left( {\ln 2 - \ln 6 + 2 + \ln 5} \right)\\
= \ln 3 - \ln 1 + 1 + 2\\
= 3 + \ln 3
\end{array}\)
