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

Xét tích phân: \(I = \int\limits_0^1 {\dfrac{{f\left( x \right)}}{{{{\left( {x + 1} \right)}^2}}}dx} \).

Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = \dfrac{{dx}}{{{{\left( {x + 1} \right)}^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v =  - \dfrac{1}{{x + 1}} + \dfrac{1}{2} = \dfrac{{x - 1}}{{2\left( {x + 1} \right)}}\end{array} \right.\)

Khi đó ta có:

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

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

\(\begin{array}{l} \Leftrightarrow \int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx}  + 2k\int\limits_0^1 {\dfrac{{x - 1}}{{x + 1}}f'\left( x \right)dx}  + {k^2}\int\limits_0^1 {{{\left( {\dfrac{{x - 1}}{{x + 1}}} \right)}^2}dx}  = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\int\limits_0^1 {{{\left( {1 - \dfrac{2}{{x + 1}}} \right)}^2}dx}  = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\int\limits_0^1 {\left( {1 - \dfrac{4}{{x + 1}} + \dfrac{4}{{{{\left( {x + 1} \right)}^2}}}} \right)dx}  = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\left. {\left( {x - 4\ln \left| {x + 1} \right| - \dfrac{4}{{x + 1}}} \right)} \right|_0^1 = 0\\ \Leftrightarrow 12 - 16\ln 2 + 2k.\left( {6 - 8\ln 2} \right) + {k^2}\left( {1 - 4\ln 2 - 2 + 4} \right) = 0\\ \Leftrightarrow \left( {3 - 4\ln 2} \right){k^2} - 4\left( {3 - 4\ln 2} \right)k + 4\left( {3 - 4\ln 2} \right) = 0\\ \Leftrightarrow {k^2} - 4k + 4 = 0 \Leftrightarrow {\left( {k - 2} \right)^2} = 0 \Leftrightarrow k = 2\end{array}\)

Khi đó ta có \(\int\limits_0^1 {{{\left( {f'\left( x \right) - 2.\dfrac{{x - 1}}{{x + 1}}} \right)}^2}dx}  = 0 \Leftrightarrow f'\left( x \right) = 2.\dfrac{{x - 1}}{{x + 1}}\).

\( \Rightarrow f\left( x \right) = \int {f'\left( x \right)dx}  = 2\int {\dfrac{{x - 1}}{{x + 1}}dx} \) \( = 2\int {\left( {1 - \dfrac{2}{{x + 1}}} \right)dx}  = 2\left( {x - 2\ln \left| {x + 1} \right|} \right) + C\).

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

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

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

Ta có: \(J = \int\limits_0^1 {\ln \left| {x + 1} \right|dx}  = \int\limits_0^1 {\ln \left( {x + 1} \right)dx} \).

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{{x + 1}}dx\\v = x + 1\end{array} \right.\)

\(\begin{array}{l} \Rightarrow J = \left. {\left( {x + 1} \right)\ln \left( {x + 1} \right)} \right|_0^1 - \int\limits_0^1 {dx} \\ \Rightarrow J = 2\ln 2 - 1.\ln 1 - \left. x \right|_0^1\\ \Rightarrow J = 2\ln 2 - 1\end{array}\)

Vậy \(\int\limits_0^1 {f\left( x \right)dx}  = 3 - 4\left( {2\ln 2 - 1} \right) = 7 - 8\ln 2\).