Differentiate $y = \ln(x^2 + 1)$ with respect to $x$.

For $y = \ln u$, $\dfrac{dy}{dx} = \dfrac{1}{u} \cdot \dfrac{du}{dx}$. With $u = x^2 + 1$, $\dfrac{du}{dx} = 2x$, so the derivative is $\dfrac{2x}{x^2 + 1}$. Forgetting the chain-rule factor $2x$ gives $\dfrac{1}{x^2+1}$.