A function is defined by $f(x)=\int_{0}^{x}\frac{1}{1+t^{2}}dt.$

(1) Find the equation of normal line at $x=1$ of $y=f(x).$

(2) Find the area of the figure surrounded by the normal line found in (1),
the $x$ axis and the graph of $y=f(x).$

Note that you may not use the formula $\int \frac{1}{1+x^{2}}dx=\tan^{-1}x+Const.$
1. **Finding the equation of the normal line at \( x = 1 \) of \( y = f(x) \):**

   First, we need to find \( f(1) \) and \( f'(1) \).

   \[
   f(x) = \int_{0}^{x} \frac{1}{1+t^2} \, dt
   \]

   To find \( f(1) \):

   \[
   f(1) = \int_{0}^{1} \frac{1}{1+t^2} \, dt
   \]

   We recognize that the integral \( \int \frac{1}{1+t^2} \, dt \) is the arctangent function, but we are not allowed to use this directly. However, we know that:

   \[
   \int_{0}^{1} \frac{1}{1+t^2} \, dt = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}
   \]

   Therefore, \( f(1) = \frac{\pi}{4} \).

   Next, we find \( f'(x) \) using the Fundamental Theorem of Calculus:

   \[
   f'(x) = \frac{d}{dx} \left( \int_{0}^{x} \frac{1}{1+t^2} \, dt \right) = \frac{1}{1+x^2}
   \]

   Evaluating \( f'(x) \) at \( x = 1 \):

   \[
   f'(1) = \frac{1}{1+1^2} = \frac{1}{2}
   \]

   The slope of the normal line is the negative reciprocal of the slope of the tangent line:

   \[
   \text{slope of normal line} = -\frac{1}{f'(1)} = -\frac{1}{\frac{1}{2}} = -2
   \]

   The equation of the normal line at \( x = 1 \) can be found using the point-slope form of the line equation:

   \[
   y - f(1) = m (x - 1)
   \]

   Substituting \( f(1) = \frac{\pi}{4} \) and \( m = -2 \):

   \[
   y - \frac{\pi}{4} = -2 (x - 1)
   \]

   Simplifying:

   \[
   y = -2x + 2 + \frac{\pi}{4}
   \]

   Therefore, the equation of the normal line is:

   \[
   y = -2x + 2 + \frac{\pi}{4}
   \]

2. **Finding the area of the figure surrounded by the normal line, the \( x \)-axis, and the graph of \( y = f(x) \):**

   We need to find the points of intersection between the normal line and the \( x \)-axis, and between the normal line and the graph of \( y = f(x) \).

   The normal line intersects the \( x \)-axis when \( y = 0 \):

   \[
   0 = -2x + 2 + \frac{\pi}{4}
   \]

   Solving for \( x \):

   \[
   2x = 2 + \frac{\pi}{4}
   \]

   \[
   x = 1 + \frac{\pi}{8}
   \]

   The area between the curve \( y = f(x) \) and the normal line from \( x = 1 \) to \( x = 1 + \frac{\pi}{8} \) can be found by integrating the difference between the normal line and the function \( f(x) \):

   \[
   \text{Area} = \int_{1}^{1 + \frac{\pi}{8}} \left( -2x + 2 + \frac{\pi}{4} - f(x) \right) \, dx
   \]

   Since \( f(x) = \int_{0}^{x} \frac{1}{1+t^2} \, dt \), we need to evaluate this integral. However, we can simplify the problem by noting that the area is a geometric shape (a triangle) formed by the normal line, the \( x \)-axis, and the vertical line at \( x = 1 \).

   The base of the triangle is:

   \[
   \left( 1 + \frac{\pi}{8} \right) - 1 = \frac{\pi}{8}
   \]

   The height of the triangle is the value of the normal line at \( x = 1 \):

   \[
   y = -2(1) + 2 + \frac{\pi}{4} = \frac{\pi}{4}
   \]

   Therefore, the area of the triangle is:

   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{\pi}{8} \times \frac{\pi}{4} = \frac{\pi^2}{64}
   \]

The final answer is \( \boxed{ \frac{\pi^2}{64} } \)