Evaluate the definite integrals`int_0^1 (dx)/(1 +x^2)`

Options

A. `pi/4`
B. `pi/2`
C. `pi/3`
D. 0

Explanation

To evaluate the definite integral

\[ \int_0^1 \frac{dx}{1+x^2}, \]

we can follow these steps:

### Step 1: Identify the Integral
We need to integrate the function \( \frac{1}{1+x^2} \) from 0 to 1.

### Step 2: Recall the Formula for Integration
The integral of the function \( \frac{1}{a^2 + x^2} \) is given by:

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

In our case, \( a = 1 \), so we have:

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

### Step 3: Apply the Limits of Integration
Now we will evaluate the definite integral from 0 to 1:

\[ \int_0^1 \frac{dx}{1+x^2} = \left[ \tan^{-1}(x) \right]_0^1. \]

### Step 4: Calculate the Values at the Limits
Now we substitute the upper and lower limits into the antiderivative:

1. For the upper limit \( x = 1 \):

\[ \tan^{-1}(1) = \frac{\pi}{4}. \]

2. For the lower limit \( x = 0 \):

\[ \tan^{-1}(0) = 0. \]

### Step 5: Subtract the Values
Now we subtract the value at the lower limit from the value at the upper limit:

\[ \int_0^1 \frac{dx}{1+x^2} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Final Answer
Thus, the value of the definite integral is

\[ \frac{\pi}{4}. \]

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