To solve this problem, we need to find the equation of the tangent line to $F(x)$ at $x = 1$. Let us follow these steps:

Step 1: Differentiate $F(x)$

The function is given as: $F(x) = x^2 + \int_{\tan(x-1)}^{0} \ln(1+t) \, dt$

To compute $F'(x)$, we differentiate both terms:

1. Differentiate $x^2$: $\frac{d}{dx}[x^2] = 2x$

2. Differentiate the integral term: The integral has variable bounds, so we apply the Leibniz rule: $\frac{d}{dx} \int_{g_1(x)}^{g_2(x)} f(t) \, dt = f(g_2(x)) g_2'(x) - f(g_1(x)) g_1'(x)$

   Here, $f(t) = \ln(1+t)$, $g_1(x) = \tan(x-1)$, and $g_2(x) = 0$. Since $g_2(x) = 0$, $g_2'(x) = 0$, and the term involving $g_2(x)$ vanishes. Therefore: $\frac{d}{dx} \int_{\tan(x-1)}^{0} \ln(1+t) \, dt = -\ln(1+\tan(x-1)) \cdot \frac{d}{dx}[\tan(x-1)]$

   The derivative of $\tan(x-1)$ is: $\frac{d}{dx}[\tan(x-1)] = \sec^2(x-1)$

   So, the derivative of the integral is: $-\ln(1+\tan(x-1)) \cdot \sec^2(x-1)$

Combining these results, we find: $F'(x) = 2x - \ln(1+\tan(x-1)) \cdot \sec^2(x-1)$

Step 2: Evaluate $F(1)$ and $F'(1)$

To find the tangent line, we need both the value of the function and its slope at $x = 1$.

1. Evaluate $F(1)$: $F(1) = (1)^2 + \int_{\tan(1-1)}^{0} \ln(1+t) \, dt = 1 + \int_{0}^{0} \ln(1+t) \, dt = 1$

2. Evaluate $F'(1)$: Substitute $x = 1$ into $F'(x)$:

   • $\tan(1-1) = \tan(0) = 0$
   • $\ln(1+\tan(0)) = \ln(1+0) = \ln(1) = 0$

   Thus, the second term vanishes: $F'(1) = 2(1) - 0 = 2$

Step 3: Write the Equation of the Tangent Line

The equation of the tangent line is: $y - F(1) = F'(1)(x-1)$

Substitute $F(1) = 1$ and $F'(1) = 2$: $y - 1 = 2(x-1)$

Simplify: $y = 2x - 1$

Final Answer:

The equation of the tangent line is: $y = 2x - 1$

Would you like additional explanation or related problems? Here are some related questions:

1. Can you explain why the term involving $g_2(x) = 0$ vanishes?
2. What happens if we change the upper limit of the integral to a nonzero function?
3. How would the tangent line change if we shifted the value of $x$ in the derivative evaluation?
4. Can we plot the graph of $F(x)$ near $x = 1$ to visualize the tangent line?
5. How does the integral contribute to the overall shape of $F(x)$?

Tip: Always check for variable bounds in integrals when differentiating; the Leibniz rule is essential for such cases.