Choose the correct answer `int1sqrt(3)(dx)/(1+x^2)` equals (A) `pi/3` (B) `(2pi)/3` (C) `pi/6` (D) `pi/(12)`

Explanation

To solve the integral \( I = \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} \), we can follow these steps:

### Step 1: Identify the integral
We start with the integral:
\[ I = \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} \]

### Step 2: Use the known integral formula
We know that:
\[ \int \frac{dx}{1+x^2} = \tan^{-1}(x) + C \]
Thus, we can rewrite our integral using this formula:
\[ I = \left[ \tan^{-1}(x) \right]_{1}^{\sqrt{3}} \]

### Step 3: Apply the limits
Now we will evaluate the integral by applying the limits:
\[ I = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \]

### Step 4: Calculate the values of the arctangent
We know that:
- \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \) (since \( \tan(\frac{\pi}{3}) = \sqrt{3} \))
- \( \tan^{-1}(1) = \frac{\pi}{4} \) (since \( \tan(\frac{\pi}{4}) = 1 \))

Thus, we can substitute these values into our expression:
\[ I = \frac{\pi}{3} - \frac{\pi}{4} \]

### Step 5: Simplify the expression
To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12:
\[ I = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4\pi - 3\pi)}{12} = \frac{\pi}{12} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{\pi}{12} \]

### Conclusion
The correct answer is (D) \( \frac{\pi}{12} \).

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