Step 1: Set a helper angle.
Let $\theta=\cos^{-1}x$, so $x=\cos\theta$. Since $x=\dfrac{n}{n^2+1}$ is a small positive number, $\theta$ lies in the first quadrant near $\dfrac{\pi}{2}$.
Step 2: Rewrite the second inverse cosine.
Note $2x^2-1=2\cos^2\theta-1=\cos2\theta$ by the double-angle rule.
Step 3: Simplify that term.
So $\cos^{-1}(2x^2-1)=\cos^{-1}(\cos2\theta)$.
Step 4: Add the two pieces.
The expression becomes $2\theta+\cos^{-1}(\cos2\theta)$.
Step 5: Handle the principal value.
Because $x$ is small, $\theta$ is just above $\dfrac{\pi}{2}$, so $2\theta$ is just above $\pi$. Then $\cos^{-1}(\cos2\theta)=2\pi-2\theta$, since the angle must land back in $[0,\pi]$.
Step 6: Combine and finish.
Adding, $2\theta+(2\pi-2\theta)=2\pi$. \[ \boxed{2\pi} \]