Cho hàm số y = f(x) liên tục và có đạo hàm trên (-căn bậc hai của 2; căn bậc hai của hai)

Ta có $f\text{'}\left(x\right)+x\left(e^{f\left(x\right)}+2\right)+\frac{x}{e^{f\left(x\right)}}=0$

$\Rightarrow f\text{'}\left(x\right).e^{f\left(x\right)}+x\left(e^{f\left(x\right)}+2\right).e^{f\left(x\right)}+x=0$

$\iff f\text{'}\left(x\right).e^{f\left(x\right)}+x\left[\left(e^{f\left(x\right)}+2\right).e^{f\left(x\right)}+1\right]=0$

$\iff {\left[e^{f\left(x\right)}\right]}^{\text{'}}+x{\left(e^{f\left(x\right)}+1\right)}^2=0$

$\iff x=-\frac{{\left[e^{f\left(x\right)}\right]}^{\text{'}}}{{\left(e^{f\left(x\right)}+1\right)}^2}$

Lấy nguyên hàm hai vế ta được:

$\underset{}{\overset{}{\int }}xdx=\underset{}{\overset{}{\int }}-\frac{{\left[e^{f\left(x\right)}\right]}^{\text{'}}}{{\left(e^{f\left(x\right)}+1\right)}^2}dx=\underset{}{\overset{}{\int }}-\frac{{\left[e^{f\left(x\right)}+1\right]}^{\text{'}}}{{\left(e^{f\left(x\right)}+1\right)}^2}dx\iff \frac{x^2}{2}+C=\frac{1}{e^{f\left(x\right)}+1}$

Mà $f\left(1\right)=0\Rightarrow \frac{1}{2}+C=\frac{1}{e^0+1}\iff C=0$

Suy ra $\frac{x^2}{2}=\frac{1}{e^{f\left(x\right)}+1}\Rightarrow e^{f\left(x\right)}+1=\frac{2}{x^2}\iff e^{f\left(x\right)}=\frac{2}{x^2}-1\Rightarrow f\left(x\right)=\ln \left(\frac{2}{x^2}-1\right).$

Vậy $f\left(\frac{1}{2}\right)=\ln \left(\frac{2}{\frac{1}{4}}-1\right)=\ln 7.$

Chọn A.