Cho hàm số \(f\left( x \right)\) có đạo hàm liên tục trên \(\mathbb{R}\) thỏa mãn \(\int\limits_0^2 {\frac{{f'\left( x \right)}}{{x + 2}}} \;{\rm{d}}x = 3\) và \(f\left( 2 \right) - 2f\left( 0 \right) =  - 4.\) Tích phân \(\int\limits_0^1 {\frac{{f\left( {2x} \right)}}{{{{\left( {x + 1} \right)}^2}}}} \;{\rm{d}}x\) bằng

Ta có \(K = \int\limits_0^2 {\frac{{f'\left( x \right)}}{{x + 2}}} \;{\rm{d}}x = \int\limits_0^2 {\frac{{d\left( {f\left( x \right)} \right)}}{{x + 2}}}  = \left. {\left( {\frac{1}{{x + 2}} \cdot f\left( x \right)} \right)} \right|_0^2 - \int\limits_0^2 {f\left( x \right) \cdot d\left( {\frac{1}{{x + 2}}} \right)} \)

\( = \frac{1}{4} \cdot f\left( 2 \right) - \frac{1}{2} \cdot f\left( 0 \right) + \int_0^2 f \left( x \right) \cdot \frac{{dx}}{{{{\left( {x + 2} \right)}^2}}} = \frac{1}{4}\left[ {f\left( 2 \right) - 2f\left( 0 \right)} \right] + \int\limits_0^2 {\frac{{f\left( x \right)}}{{{{\left( {x + 2} \right)}^2}}}} \;{\rm{d}}x\)

\( = \frac{1}{4} \cdot \left( { - 4} \right) + \int\limits_0^2 {\frac{{f\left( x \right)}}{{{{\left( {x + 2} \right)}^2}}}} \;{\rm{d}}x =  - 1 + \int\limits_0^2 {\frac{{f\left( x \right)}}{{{{\left( {x + 2} \right)}^2}}}} \;{\rm{d}}x = 3\)\( \Rightarrow \int\limits_0^2 {\frac{{f\left( x \right)}}{{{{\left( {x + 2} \right)}^2}}}} \;{\rm{d}}x = 4.\)

Ta cần tính: \(I = \int\limits_0^1 {\frac{{f\left( {2x} \right)}}{{{{\left( {x + 1} \right)}^2}}}} \;{\rm{d}}x.\) Đặt \(t = 2x \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x = \frac{t}{2}}\\{dt = 2dx}\end{array}} \right..\)

Đổi cận: \(\left\{ {\begin{array}{*{20}{l}}{x = 0 \Rightarrow t = 0}\\{x = 1 \Rightarrow t = 2}\end{array}} \right..\)

\[ \Rightarrow I = \int\limits_0^2 {\frac{{f\left( t \right)}}{{{{\left( {\frac{t}{2} + 1} \right)}^2}}} \cdot \frac{{dt}}{2}}  = \int\limits_0^2 {\frac{{4f\left( t \right)}}{{{{\left( {t + 2} \right)}^2}}} \cdot \frac{{dt}}{2}}  = 2 \cdot \int\limits_0^2 {\frac{{f\left( t \right)}}{{{{\left( {t + 2} \right)}^2}}}dt}  = 2 \cdot 4 = 8\].

Chọn B