Find the second derivative of y = (1+x²) arctan x

Find the second derivative of $y = \bigl(1+x^{2}\bigr)\arctan x$.

Step 1: First derivative

Product rule. The factor $1+x^{2}$ is exactly the denominator of $\bigl(\arctan x\bigr)'$, so the second term reduces to $1$: $y' = 2x\arctan x+\bigl(1+x^{2}\bigr)\cdot\frac{1}{1+x^{2}} = 2x\arctan x+1$.

Step 2: Second derivative

The constant drops, and the product rule applies once more: $y'' = 2\arctan x+2x\cdot\frac{1}{1+x^{2}} = 2\arctan x+\frac{2x}{1+x^{2}}$.

Step 3: The origin is an inflection point

Both terms of $y''$ are odd functions, so $y''$ is odd and $y''(0) = 0$. To decide whether this is a genuine inflection, differentiate once more: \[y''' = \frac{2}{1+x^{2}}+\frac{2\bigl(1-x^{2}\bigr)}{\bigl(1+x^{2}\bigr)^{2}}, \qquad y'''(0) = 2+2 = 4 \neq 0.\]

Since $y''(0) = 0$ and $y'''(0)\neq0$, the second derivative changes sign at the origin, and $x = 0$ is an inflection point. The curve is concave for $x\lt 0$ and convex for $x>0$.

Step 4: Verification

\[y''(-1) = -\frac{\pi}{2}-1 \approx -2.5708 \lt 0, \qquad y''(1) = \frac{\pi}{2}+1 \approx 2.5708 >0 \;\;\checkmark\]

\[\text{The sign change confirms the inflection at } x = 0 \;\;\checkmark\]

\[x\to+\infty:\ y'' \to 2\cdot\frac{\pi}{2}+0 = \pi, \text{ so the curvature does not die away} \;\;\checkmark\]