Cho hàm số $f\left( x \right)$ có $f\left( 2 \right) = 1$ và $3f\left( x \right) = \left( {1 - 3x} \right)f'\left( x \ri?
Cho hàm số \(f\left( x \right)\) có \(f\left( 2 \right) = 1\) và \(3f\left( x \right) = \left( {1 - 3x} \right)f'\left( x \right) + \dfrac{x}{{\sqrt {{x^2} + 5} }},\forall x \in \left( {\dfrac{1}{3}; + \,\infty } \right)\). Khi đó \(\int\limits_1^2 {\dfrac{{f\left( x \right)}}{{\sqrt {{x^2} + 5} + 2}}{\rm{d}}x} \) bằng
A. \(\ln \dfrac{5}{2}.\)
B. \(\dfrac{1}{3}\ln \dfrac{2}{5}.\)
C. \(\dfrac{1}{3}\ln 10.\)
D. \(\dfrac{1}{3}\ln \dfrac{5}{2}.\)
Đáp án D
Chọn D\(\begin{array}{l} 3f\left( x \right) = \left( {1 - 3x} \right)f'\left( x \right) + \dfrac{x}{{\sqrt {{x^2} + 5} }} \Leftrightarrow \left( {1 - 3x} \right)f'\left( x \right) - 3f\left( x \right) = - \dfrac{x}{{\sqrt {{x^2} + 5} }}\\ \Leftrightarrow {\left[ {\left( {1 - 3x} \right)f\left( x \right)} \right]^\prime } = - \dfrac{x}{{\sqrt {{x^2} + 5} }}\\ \Rightarrow \left( {1 - 3x} \right)f\left( x \right) = - \,\sqrt {{x^2} + 5} + C\\ f\left( 2 \right) = 1 \Rightarrow - 5.1 = - 3 + C \Rightarrow C = - 2\\ \Rightarrow f\left( x \right) = \dfrac{{ - \,\sqrt {{x^2} + 5} - 2}}{{1 - 3x}} = \dfrac{{\sqrt {{x^2} + 5} + 2}}{{3x - 1}}\\ \int\limits_1^2 {\dfrac{{f\left( x \right)}}{{\sqrt {{x^2} + 5} + 2}}{\rm{d}}x} = \int\limits_1^2 {\dfrac{1}{{3x - 1}}{\rm{d}}x = } \left. {\dfrac{1}{3}\ln \left| {3x - 1} \right|} \right|_1^2 = \dfrac{1}{3}\ln \dfrac{5}{2} \end{array}\)