Cho hàm số $\Large y=f(x)$ liên tục trên $\Large \mathbb{R}$ và có đạo

Cho hàm số $\Large y=f(x)$ liên tục trên $\Large \mathbb{R}$ và có đạo hàm là $\Large {f}'(x)$ . Biết rằng $\Large {{f}^{2}}(2)=6+8{{f}^{2}}(1)$, $\Large \int\limits_{1}^{2}{\dfrac{2x+1}{x+{{f}^{2}}(x)}dx=\dfrac{11}{16}}$ . Tính $\Large I=\int\limits_{1}^{2}{\dfrac{f(x)+{f}'(x)}{x+{{f}^{2}}(x)}.f(x)dx}$

• A.

  $\Large I=\dfrac{21}{16}+3\ln 2$

• B.

  $\Large I=\dfrac{21}{32}+\dfrac{3}{2}\ln 2$

• C.

  $\Large I=\dfrac{21}{32}+\ln 2$

• D.

  $\Large I=\dfrac{21}{16}-\dfrac{3}{2}\ln 2$

Ta có: $\Large {{f}^{2}}(2)=6+8{{f}^{2}}(1)\Leftrightarrow 2+{{f}^{2}}(2)=8+8{{f}^{2}}(1)\Leftrightarrow \dfrac{2+{{f}^{2}}(2)}{1+{{f}^{2}}(1)}=8$

$\Large I=\int\limits_{1}^{2}{\dfrac{f(x)+{f}'(x)}{x+{{f}^{2}}(x)}f(x)dx=\int\limits_{1}^{2}{\dfrac{{{f}^{2}}(x)+{f}'(x).f(x)}{x+{{f}^{2}}(x)}}dx}$

$\Large =\int\limits_{1}^{2}{\dfrac{\left( {{f}^{2}}(x)+x \right)+\left( {f}'(x).f(x)-x \right)}{x+{{f}^{2}}(x)}dx=\int\limits_{1}^{2}{\left[ 1+\dfrac{1}{2}.\dfrac{1+2{f}'(x).f(x)-2x-1}{x+{{f}^{2}}(x)} \right]dx}}$

$\Large =\int\limits_{1}^{2}{\left[ 1+\dfrac{1}{2}.\dfrac{1+2{f}'(x).f(x)}{x+{{f}^{2}}(x)}-\dfrac{1}{2}.\dfrac{2x+1}{x+{{f}^{2}}(x)} \right]dx}$

$\Large =\int\limits_{1}^{2}{dx+\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{1+2{f}'(x).f(x)}{x+{{f}^{2}}(x)}dx-\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{2x+1}{x+{{f}^{2}}(x)}dx}}}$

$\Large =\int\limits_{1}^{2}{dx+\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{d(x+{{f}^{2}}(x))}{x+{{f}^{2}}(x)}-\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{2x+1}{x+{{f}^{2}}(x)}dx}}}$

$\Large =x\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.$ $\Large +\dfrac{1}{2}\ln \left| x+{{f}^{2}}(x) \right|\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.$ $\Large -\dfrac{1}{2}.\dfrac{11}{6}$

$\Large =(2-1)+\dfrac{1}{2}(\ln \left| 2+{{f}^{2}}(2) \right|-\ln \left| 1+{{f}^{2}}(1) \right|)-\dfrac{11}{32}$

$\Large = \dfrac{21}{32}+\dfrac{1}{2}\ln \left| \dfrac{2+{{f}^{2}}(2)}{1+{{f}^{2}}(1)} \right|$

$\Large = \dfrac{21}{32}+\dfrac{1}{2}\ln 8=\dfrac{21}{32}+\dfrac{3}{2}\ln 2$

Chọn đáp án B