Evaluate: `int_1^(sqrt3) 1/(1+x^2) dx`

Answer

To evaluate the integral

\[ \int_1^{\sqrt{3}} \frac{1}{1+x^2} \, dx, \]

we can follow these steps:

### Step 1: Identify the Integral
The integral we need to evaluate is

\[ \int \frac{1}{1+x^2} \, dx. \]

### Step 2: Use the Standard Integral Formula
We know from calculus that the integral of \(\frac{1}{1+x^2}\) is

\[ \arctan(x) + C. \]

### Step 3: Apply the Limits
Now, we will evaluate the definite integral from \(x = 1\) to \(x = \sqrt{3}\):

\[ \int_1^{\sqrt{3}} \frac{1}{1+x^2} \, dx = \left[ \arctan(x) \right]_1^{\sqrt{3}}. \]

### Step 4: Substitute the Limits
Now we substitute the upper and lower limits into the antiderivative:

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

### Step 5: Evaluate the Arctangent Values
We know that:

- \(\arctan(\sqrt{3}) = \frac{\pi}{3}\) (since \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\)),
- \(\arctan(1) = \frac{\pi}{4}\) (since \(\tan\left(\frac{\pi}{4}\right) = 1\)).

### Step 6: Calculate the Result
Now we can compute:

\[ \arctan(\sqrt{3}) - \arctan(1) = \frac{\pi}{3} - \frac{\pi}{4}. \]

To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12:

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

Thus,

\[ \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{1\pi}{12} = \frac{\pi}{12}. \]

### Final Answer
The value of the integral is

\[ \int_1^{\sqrt{3}} \frac{1}{1+x^2} \, dx = \frac{\pi}{12}. \]

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