Question:

Step 1: Apply the Fundamental Theorem of Calculus to find the first derivative. Since $f(x) = \int_0^x t\ln(1+t^2)\, dt$, by the Fundamental Theorem of Calculus (Leibniz rule), we differentiate with respect to $x$ by substituting $x$ into the integrand: $$f'(x) = x\ln(1+x^2)$$ Step 2: Differentiate $f'(x)$ to find the second derivative. We need to find $f''(x)$ by differentiating $f'(x) = x\ln(1+x^2)$ using the product rule: $$f''(x) = \frac{d}{dx}\left[x\ln(1+x^2)\right]$$ Applying the product rule where $u = x$ and $v = \ln(1+x^2)$: $$f''(x) = 1 \cdot \ln(1+x^2) + x \cdot \frac{d}{dx}[\ln(1+x^2)]$$ For the derivative of $\ln(1+x^2)$, we use the chain rule: $$\frac{d}{dx}[\ln(1+x^2)] = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}$$ Therefore: $$f''(x) = \ln(1+x^2) + x \cdot \frac{2x}{1+x^2} = \ln(1+x^2) + \frac{2x^2}{1+x^2}$$ Step 3: Evaluate $f''(x)$ at $x = 0$. Substituting $x = 0$ into the expression for $f''(x)$: $$f''(0) = \ln(1+0^2) + \frac{2(0)^2}{1+0^2}$$ $$f''(0) = \ln(1) + \frac{0}{1}$$ $$f''(0) = 0 + 0 = 0$$ **Final Answer:** $f''(0) = 0$ The answer is **Option 1: (a) 0**