\({\rm{Ta c\'o : }}f(x) = \int {\tan } xdx = \int {\frac{{\sin xdx}}{{\cos x}}} = - \int {\frac{{d(\cos x)}}{{\cos x}}} = - \ln |\cos x| + C\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f(x) = - \ln (\cos x) + {C_1}\quad khix \in \left[ { - \frac{\pi }{4};\frac{\pi }{2}} \right)}\\{f(x) = - \ln ( - \cos x) + {C_2}\;khix \in \left( {\frac{\pi }{2};\frac{{5\pi }}{4}} \right]}\end{array}} \right.\)

\({\rm{M\`a }}\,\,f(0) = 0,f(\pi ) = 1 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{ - ln 1 + {C_1} = 0}\\{ - ln 1 + {C_2} = 1}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{C_1} = 0}\\{{C_2} = 1}\end{array}} \right.} \right.\).

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f(x) = - \ln (\cos x)\,\,\,\,\,khi\,\,x \in \left[ { - \frac{\pi }{4};\frac{\pi }{2}} \right)}\\{f(x) = - \ln ( - \cos x) + 1\quad khi\,\,x \in \left( {\frac{\pi }{2};\frac{{5\pi }}{4}} \right]}\end{array}} \right.\) .

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f\left( {\frac{\pi }{4}} \right) = - \ln \left( {\cos \frac{\pi }{4}} \right) = - \ln \frac{{\sqrt 2 }}{2} = - \frac{1}{2}\ln 2 + \ln 2 = \frac{1}{2}\ln 2}\\{f\left( {\frac{{2\pi }}{3}} \right) = - \ln \left( { - \cos \frac{{2\pi }}{3}} \right) + 1 = - \ln \frac{1}{2} + 1 = \ln 2 + 1}\end{array}} \right.\)

\( \Rightarrow \frac{{f\left( {\frac{{2\pi }}{3}} \right)}}{{f\left( {\frac{\pi }{4}} \right)}} = \frac{{2(\ln 2 + 1)}}{{\ln 2}} = 2\left( {1 + {{\log }_2}e} \right)\). Chọn A.