Choose the correct answer of the given question
`int dx/(1+x^2) = ___ +c`

Options

A. `tan^-1x`
B. `sin^-1x`
C. `cos^-1x`
D. `cot^-1x`

Answer

To solve the integral \( \int \frac{dx}{1+x^2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Integral**: We start with the integral
\[ I = \int \frac{dx}{1+x^2}. \]

2. **Recognize the Standard Form**: We know that the integral
\[ \int \frac{dx}{1+x^2} = \tan^{-1}(x) + C, \]
where \( C \) is the constant of integration.

3. **Substitution (if needed)**: Although we can directly use the standard form, we can also verify this by using a substitution. Let
\[ x = \tan(\theta). \]
Then, the differential \( dx \) is given by
\[ dx = \sec^2(\theta) d\theta. \]

4. **Change of Variables**: Substitute \( x = \tan(\theta) \) into the integral:
\[ I = \int \frac{\sec^2(\theta) d\theta}{1+\tan^2(\theta)}. \]

5. **Simplify the Denominator**: Using the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \), we can simplify the integral:
\[ I = \int \frac{\sec^2(\theta) d\theta}{\sec^2(\theta)} = \int d\theta. \]

6. **Integrate**: The integral of \( d\theta \) is
\[ I = \theta + C. \]

7. **Back-Substitute**: Now, we need to revert back to \( x \). Since we set \( x = \tan(\theta) \), we have
\[ \theta = \tan^{-1}(x). \]

8. **Final Result**: Therefore, we can write the integral as
\[ I = \tan^{-1}(x) + C. \]

### Conclusion:
The final answer is
\[ \int \frac{dx}{1+x^2} = \tan^{-1}(x) + C. \]