Choose the correct answer`int_1^sqrt(3) (dx)/(1+x^2)`equals
(A) `pi/3`
(B) `(2pi)/3`
(C) `pi/6`
(D) `pi/(12)`

Explanation

To solve the integral \( \int_1^{\sqrt{3}} \frac{dx}{1+x^2} \), we will follow these steps:

### Step 1: Identify the integral
We need to evaluate the integral:
\[ \int_1^{\sqrt{3}} \frac{dx}{1+x^2} \]

### Step 2: Find the antiderivative
The antiderivative of \( \frac{1}{1+x^2} \) is:
\[ \tan^{-1}(x) \]

### Step 3: Apply the limits of integration
Now, we will apply the limits from 1 to \( \sqrt{3} \):
\[ \left[ \tan^{-1}(x) \right]_1^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \]

### Step 4: Evaluate \( \tan^{-1}(\sqrt{3}) \) and \( \tan^{-1}(1) \)
We know:
- \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \) (since \( \tan(\frac{\pi}{3}) = \sqrt{3} \))
- \( \tan^{-1}(1) = \frac{\pi}{4} \) (since \( \tan(\frac{\pi}{4}) = 1 \))

### Step 5: Substitute the values
Now substituting these values back into the expression:
\[ \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) = \frac{\pi}{3} - \frac{\pi}{4} \]

### Step 6: Find a common denominator
To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12:
\[ \frac{\pi}{3} = \frac{4\pi}{12}, \quad \frac{\pi}{4} = \frac{3\pi}{12} \]

### Step 7: Perform the subtraction
Now we can subtract:
\[ \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4\pi - 3\pi)}{12} = \frac{\pi}{12} \]

### Conclusion
Thus, the value of the integral is:
\[ \int_1^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{12} \]

### Final Answer
The correct answer is \( \frac{\pi}{12} \), which corresponds to option (D).

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