Chọn phương án A.

Đặt $m=\displaystyle\int\limits_{0}^{1}xf(x)\mathrm{\,d}x$ ta có $f(x)=\mathrm{e}^x+m$. Khi đó $$\begin{aligned}
m&=\displaystyle\int\limits_{0}^{1}xf(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}x\left(\mathrm{e}^x+m\right)\mathrm{\,d}x\\
&=\displaystyle\int\limits_{0}^{1}\left(x\mathrm{e}^x+mx\right)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}x\mathrm{e}^x\mathrm{\,d}x+\displaystyle\int\limits_{0}^{1}mx\mathrm{\,d}x\\
&=I+\dfrac{mx^2}{2}\bigg|_0^1=I+\dfrac{m}{2}.
\end{aligned}$$
Đặt $\begin{cases}
u=x\\
v'=\mathrm{e}^x
\end{cases}\Rightarrow\begin{cases}
u'=1\\
v=\mathrm{e}^x
\end{cases}$
Khi đó $I=x\mathrm{e}^x\bigg|_0^1-\displaystyle\int\limits_{0}^{1}\mathrm{e}^x\mathrm{\,d}x=\mathrm{e}-(\mathrm{e}-1)=1$.

Vậy $m=1+\dfrac{m}{2}\Leftrightarrow\dfrac{m}{2}=1\Leftrightarrow m=2$.

Do đó $f(x)=\mathrm{e}^x+2$. Suy ra $f\left(\ln5620\right)=\mathrm{e}^{\ln5620}+2=5622$.