Chọn phương án C.

Ta có $I=\displaystyle\int\limits_{0}^{a}\dfrac{x\left(x^2+1\right)}{\sqrt{x^2+1}}\mathrm{\,d}x=\displaystyle\int\limits_{0}^{a}x\sqrt{x^2+1}\mathrm{\,d}x$.

Đặt $u=\sqrt{x^2+1}\Leftrightarrow u^2=x^2+1$.
Khi đó $2u\mathrm{d}u=2x\mathrm{d}x\Leftrightarrow u\mathrm{d}u=x\mathrm{d}x$.

• $x=a\Rightarrow u=\sqrt{a^2+1}$.
• $x=0\Rightarrow u=1$.

Vậy $I=\displaystyle\int\limits_{1}^{\sqrt{a^2+1}}u^2\mathrm{\,d}u=\dfrac{u^3}{3}\bigg|_{1}^{\sqrt{a^2+1}}=\dfrac{1}{3}\left[\left(a^2+1\right)\sqrt{a^2+1}-1\right]$.