Chọn B

Do $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=1\iff \underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=1\iff f\text{'}\left(0\right)=1$.

Ta có: $f\text{'}\text{'}\left(x\right)+{\left[f\text{'}\left(x\right)\right]}^2+x^2=1+2xf\text{'}\left(x\right)\iff {\left[f\text{'}\left(x\right)-x\right]}^2=-\left(f\text{'}\text{'}\left(x\right)-1\right)$, (1)

Đặt $g\left(x\right)=f\text{'}\left(x\right)-x\Rightarrow g\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)-1$, nên (1) trở thành $g^2\left(x\right)=-g\text{'}\left(x\right)\Rightarrow \frac{g\text{'}\left(x\right)}{g^2\left(x\right)}=-1.$

Lấy nguyên hàm hai vế, ta được $-\frac{1}{g\left(x\right)}=-x+C\Rightarrow g\left(x\right)=\frac{1}{x-C}\Rightarrow f\text{'}\left(x\right)=x+\frac{1}{x-C}$

Cho $x=0\Rightarrow f\text{'}\left(0\right)=\frac{1}{-C}\Rightarrow C=-1$. Do đó $f\text{'}\left(x\right)=x+\frac{1}{x+1}\Rightarrow f\left(x\right)=\frac{x^2}{2}+\ln \left|x+1\right|+C_1$