`int_(0)^(pi//4) (dx)/(1+x^(2))`

Options

A. 0
B. 1
C. `tan^(-1) (pi/4)`
D. `-tan ^(-1)(pi/4)`

Answer

To solve the definite integral \(\int_{0}^{\frac{\pi}{4}} \frac{dx}{1+x^2}\), we can follow these steps:

### Step 1: Identify the integral
The integral we are working with is:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{dx}{1+x^2} \]

### Step 2: Recognize the form of the integral
The integrand \(\frac{1}{1+x^2}\) is a standard form whose integral is known:
\[ \int \frac{dx}{1+x^2} = \tan^{-1}(x) + C \]
where \(C\) is the constant of integration for indefinite integrals.

### Step 3: Apply the definite integral limits
Since we are dealing with a definite integral, we can directly apply the limits:
\[ I = \left[ \tan^{-1}(x) \right]_{0}^{\frac{\pi}{4}} \]

### Step 4: Evaluate the integral at the limits
Now we will evaluate \(\tan^{-1}(x)\) at the upper limit \(\frac{\pi}{4}\) and the lower limit \(0\):
\[ I = \tan^{-1}\left(\frac{\pi}{4}\right) - \tan^{-1}(0) \]

### Step 5: Calculate the values
We know that:
\[ \tan^{-1}(0) = 0 \]
And:
\[ \tan^{-1}\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \]
Thus, substituting these values back into our equation gives:
\[ I = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Final Answer
Therefore, the value of the definite integral is:
\[ I = \frac{\pi}{4} \]

---