Chọn phương án B.

Đặt \(u=x^2+2\Rightarrow\mathrm{d}u=2x\mathrm{d}x\).

Khi đó ta có $$\begin{aligned}
\displaystyle\int f\left(x\right)\mathrm{d}x&=\displaystyle\int\dfrac{x}{\sqrt{x^2+2}}\mathrm{d}x\\
&=\displaystyle\int\dfrac{2x}{2\sqrt{x^2+2}}\mathrm{d}x\\
&=\displaystyle\int\dfrac{1}{2\sqrt{u}}\mathrm{d}x\\
&=\sqrt{u}+C\\
&=\sqrt{x^2+2}+C.
\end{aligned}$$
Đặt \(\begin{cases}
u=x+1\\ v'=f'(x)
\end{cases}\Rightarrow\begin{cases}
u'=1\\ v=f(x)
\end{cases}\). Ta có $$\begin{aligned}
\displaystyle\int g\left(x\right)\mathrm{d}x&=\displaystyle\int\left(x+1\right)f'\left(x\right)\mathrm{d}x\\
&=\left(x+1\right)f\left(x\right)-\displaystyle\int f\left(x\right)\mathrm{d}x\\
&=\dfrac{x^2+x}{\sqrt{x^2+2}}-\sqrt{x^2+2}+C\\
&=\dfrac{x-2}{\sqrt{x^2+2}}+C.
\end{aligned}$$