Đáp án đúng: D
Giải chi tiết:Ta có:
\(\begin{array}{l}\left( {f\left( x \right)\tan x} \right)' = f'\left( x \right)\tan x + f\left( x \right).\frac{1}{{{{\cos }^2}x}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f'\left( x \right)\tan x + f\left( x \right)\left( {{{\tan }^2}x + 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f'\left( x \right)\tan x + f\left( x \right){\tan ^2}x + f\left( x \right)\\ \Rightarrow f'\left( x \right)\tan x + f\left( x \right){\tan ^2}x = \left( {f\left( x \right)\tan x} \right)' - f\left( x \right)\\ \Rightarrow I = \int\limits_0^3 {\left[ {f\left( x \right){{\tan }^2}x + f'\left( x \right)\tan x} \right]dx} \\ \Leftrightarrow I = \int\limits_0^3 {\left[ {\left( {f\left( x \right)\tan x} \right)' - f\left( x \right)} \right]dx} \\ \Leftrightarrow I = \int\limits_0^3 {\left[ {\left( {f\left( x \right)\tan x} \right)'} \right]dx} - \int\limits_0^3 {f\left( x \right)dx} \\ \Leftrightarrow I = f\left( 3 \right)\tan 3 - f\left( 0 \right)\tan 0 - 10\\ \Leftrightarrow I = \cot 3.\tan 3 - 0 - 10\\ \Leftrightarrow I = 1 - 10\\ \Leftrightarrow I = - 9\end{array}\)
Chọn D.
