A segment of length $1$ is drawn such that its endpoints lie on a unit circle, dividing the circle into two parts. Compute the area of the larger region.
1. **Identify the Geometry**:
   - Let \( O \) be the center of the unit circle.
   - Let \( X \) and \( Y \) be the endpoints of the segment of length 1.
   - Since the segment \( XY \) is of length 1 and lies on the unit circle, \( OXY \) forms an equilateral triangle with side length 1.

2. **Calculate the Area of the Equilateral Triangle**:
   - The area \( A \) of an equilateral triangle with side length \( s \) is given by:
     \[
     A = \frac{\sqrt{3}}{4} s^2
     \]
   - Here, \( s = 1 \), so the area of \( \triangle OXY \) is:
     \[
     A = \frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}
     \]

3. **Calculate the Central Angle**:
   - Since \( OXY \) is an equilateral triangle, the central angle \( \angle XOY \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians.

4. **Calculate the Area of the Sector**:
   - The area of a sector of a circle with radius \( r \) and central angle \( \theta \) is given by:
     \[
     \text{Area of sector} = \frac{1}{2} r^2 \theta
     \]
   - Here, \( r = 1 \) and \( \theta = \frac{\pi}{3} \), so the area of the sector \( OXY \) is:
     \[
     \text{Area of sector} = \frac{1}{2} \cdot 1^2 \cdot \frac{\pi}{3} = \frac{\pi}{6}
     \]

5. **Calculate the Area of the Segment**:
   - The area of the segment is the area of the sector minus the area of the triangle:
     \[
     \text{Area of segment} = \frac{\pi}{6} - \frac{\sqrt{3}}{4}
     \]

6. **Calculate the Area of the Larger Region**:
   - The total area of the circle is \( \pi \).
   - The area of the smaller region is the area of the segment:
     \[
     \text{Area of smaller region} = \frac{\pi}{6} - \frac{\sqrt{3}}{4}
     \]
   - The area of the larger region is the total area of the circle minus the area of the smaller region:
     \[
     \text{Area of larger region} = \pi - \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = \pi - \frac{\pi}{6} + \frac{\sqrt{3}}{4} = \frac{5\pi}{6} + \frac{\sqrt{3}}{4}
     \]

Conclusion:
\[
\boxed{\frac{5\pi}{6} + \frac{\sqrt{3}}{4}}
\]