Chọn phương án B.

Ta có $$\begin{aligned}
f'\left(x\right)&=\dfrac{x}{x+1-\sqrt{x+1}}\\
&=\dfrac{x}{\sqrt{x+1}\left(\sqrt{x+1}-1\right)}\\
&=\dfrac{x\left(\sqrt{x+1}+1\right)}{\sqrt{x+1}\left(\sqrt{x+1}-1\right)\left(\sqrt{x+1}+1\right)}\\
&=\dfrac{x\left(\sqrt{x+1}+1\right)}{\sqrt{x+1}\left((x+1)-1\right)}\\
&=\dfrac{x\left(\sqrt{x+1}+1\right)}{x\sqrt{x+1}}\\
&=\dfrac{\sqrt{x+1}+1}{\sqrt{x+1}}\\
&=1+\dfrac{1}{\sqrt{x+1}}.
\end{aligned}$$
Khi đó $$\begin{aligned}
f\left(x\right)&=\displaystyle\int f'(x)\mathrm{\,d}x\\
&=\displaystyle\int{\left(1+\dfrac{1}{\sqrt{x+1}}\right)}\mathrm{\,d}x\\
&=\displaystyle\int{\left(1+\dfrac{2}{2\sqrt{x+1}}\right)}\mathrm{\,d}x\\
&=x+2\sqrt{x+1}+C.
\end{aligned}$$
Mà \(f\left(3\right)=3\) nên $$3+2\sqrt{3+1}+C=3\Leftrightarrow C=-4.$$Vậy \(f(x)=x+2\sqrt{x+1}-4\).

Khi đó $$\begin{aligned}
\displaystyle\int\limits_3^8f\left(x\right)\mathrm{\,d}x&=\displaystyle\int\limits_3^8\left(x+2\sqrt{x+1}-4\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_3^8\left(x+2(x+1)^{\tfrac{1}{2}}-4\right)\mathrm{\,d}x\\
&=\left(\dfrac{x^2}{2}+\dfrac{4}{3}\sqrt{(x+1)^3}-4x\right)\bigg|_3^8\\
&=\dfrac{197}{6}.
\end{aligned}$$