`int_(1)^(sqrt(3))(dx)/(1+x^(2))` equals

Options

A. `(pi)/(3)`
B. `(2pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(12)`

Answer

To solve the integral

\[ \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}, \]

we can use the standard formula for integration:

\[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C, \]

where \( a = 1 \) in our case.

### Step-by-step Solution:

1. **Identify the formula to use**:
We recognize that the integral can be solved using the formula mentioned above. Here, \( a = 1 \).

2. **Apply the formula**:
According to the formula, we can write:

\[ \int \frac{dx}{1+x^2} = \tan^{-1}(x) + C. \]

3. **Evaluate the definite integral**:
We need to evaluate the integral from \( x = 1 \) to \( x = \sqrt{3} \):

\[ \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \left[ \tan^{-1}(x) \right]_{1}^{\sqrt{3}}. \]

4. **Substitute the limits**:
Now we substitute the upper and lower limits into the antiderivative:

\[ = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1). \]

5. **Calculate the values**:
We know from trigonometric values that:

- \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)

Therefore, we have:

\[ = \frac{\pi}{3} - \frac{\pi}{4}. \]

6. **Find a common denominator**:
To subtract these fractions, we find a common denominator, which is 12:

\[ = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}. \]

7. **Final answer**:
Thus, the value of the integral is:

\[ \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{12}. \]