Cho hàm số f(x) liên tục trên \[\left( { - \frac{1}{2};2} \right)\;\]thỏa mãn \[f\left( 0 \right) = 2\], \({\int\limits_0^1 {\left[ {f'\left( x \right)} \right]} ^2}dx = 12 - 16\ln 2,\int\limits_0^1 {\frac{{f\left( x \right)}}{{{{\left( {x + 1} \right)}^2}}}} dx = 4\ln 2 - 2\). Tính \(\int\limits_0^1 {f\left( x \right)} dx\)

Xét tích phân: \[I = \mathop \smallint \limits_0^1 \frac{{f\left( x \right)}}{{{{\left( {x + 1} \right)}^2}}}dx\]

Đặt \(\left\{ {\begin{array}{*{20}{c}}{u = f(x)}\\{dv = \frac{{dx}}{{{{\left( {x + 1} \right)}^2}}}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = f\prime (x)dx}\\{v = - \frac{1}{{x + 1}} + \frac{1}{2} = \frac{{x - 1}}{{2\left( {x + 1} \right)}}}\end{array}} \right.\)

Khi đó ta có:

\(\begin{array}{l}I = \frac{{x - 1}}{{2(x + 1)}}f\left( x \right)\left| {_0^1} \right. - \int\limits_0^1 {\frac{{x - 1}}{{2(x + 1)}}f'\left( x \right)} dx\\ \Leftrightarrow I = \frac{1}{2}f\left( 0 \right) - \frac{1}{2}\int\limits_0^1 {\frac{{x - 1}}{{x + 1}}f'\left( x \right)} dx\\ \Leftrightarrow 4ln2 - 2 = \frac{1}{2}.2 - \frac{1}{2}\int\limits_0^1 {\frac{{x - 1}}{{x + 1}}f'\left( x \right)} dx\\ \Leftrightarrow \int\limits_0^1 {\frac{{x - 1}}{{x + 1}}f'\left( x \right)} dx = 6 - 8ln2\end{array}\)

Xét\({\int\limits_0^1 {\left( {f\prime (x) + k\frac{{x - 1}}{{x + 1}}} \right)} ^2}dx = 0\)

\( \Leftrightarrow \int\limits_0^1 {{{[f\prime (x)]}^2}dx + 2k} \int\limits_0^1 {\frac{{x - 1}}{{x + 1}}} f'\left( x \right)dx + {k^2}\int\limits_0^1 {{{\left( {\frac{{x - 1}}{{x + 1}}} \right)}^2}} dx\)

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}\int\limits_0^1 {{{\left( {1 - \frac{2}{{x + 1}}} \right)}^2}} dx = 0\]

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}\int\limits_0^1 {\left( {1 - \frac{4}{{x + 1}} + \frac{4}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx = 0\]

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}\left( {x - 4ln|x + 1| - \frac{4}{{x + 1}}} \right)\left| {_0^1} \right. = 0\]

\[ \Leftrightarrow 12 - 16ln2 + 2k.(6 - 8ln2) + {k^2}(1 - 4ln2 - 2 + 4) = 0\]

\[ \Leftrightarrow (3 - 4ln2){k^2} - 4(3 - 4ln2)k + 4(3 - 4ln2) = 0\]

\[ \Leftrightarrow {k^2} - 4k + 4 = 0 \Leftrightarrow {(k - 2)^2} = 0 \Leftrightarrow k = 2\]

Khi đó ta có\[\mathop \smallint \limits_0^1 {\left( {f'\left( x \right) - 2.\frac{{x - 1}}{{x + 1}}} \right)^2}dx = 0 \Leftrightarrow f'\left( x \right) = 2.\frac{{x - 1}}{{x + 1}}\]

\[ \Rightarrow f\left( x \right) = \smallint f'\left( x \right)dx = 2\smallint \frac{{x - 1}}{{x + 1}}dx\]

\[ = 2\smallint \left( {1 - \frac{2}{{x + 1}}} \right)dx = 2\left( {x - 2\ln \left| {x + 1} \right|} \right) + C\]

Có\[f\left( 0 \right) = 2 \Rightarrow 2\left( {0 - 2\ln 1} \right) + C = 2 \Leftrightarrow C = 2\]

\[ \Rightarrow f\left( x \right) = 2\left( {x - 2\ln \left| {x + 1} \right|} \right) + 2 = 2x - 4\ln \left| {x + 1} \right| + 2\]

\( \Rightarrow \int\limits_0^1 {f(x)dx = \int\limits_0^1 {[2x - 4ln|x + 1| + 2]dx} } \)

\[ = ({x^2} + 2x)\left| {_0^1} \right. - 4\int\limits_0^1 {ln|x + 1|dx = 3 - 4J} \]

Ta có:\[J = \mathop \smallint \limits_0^1 \ln \left| {x + 1} \right|dx = \mathop \smallint \limits_0^1 \ln \left( {x + 1} \right)dx\]

Đặt\(\left\{ {\begin{array}{*{20}{c}}{u = ln(x + 1)}\\{dv = dx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = \frac{1}{{x + 1}}dx}\\{v = x + 1}\end{array}} \right.\)

\[ \Rightarrow J = (x + 1)ln(x + 1)\left| {_0^1} \right. - \int\limits_0^1 {dx} \]

\[ \Rightarrow J = 2ln2 - 1.ln1 - x\left| {_0^1} \right.\]

\[ \Rightarrow J = 2ln2 - 1\]

Vậy\[\mathop \smallint \limits_0^1 f\left( x \right)dx = 3 - 4\left( {2\ln 2 - 1} \right) = 7 - 8\ln 2\]

Đáp án cần chọn là: D