Question

\(\int \frac{dx}{1 + x^2}\)

• A. \(\tan x + \text{constant}\)
• B. \(\tan^{-1} x + \text{constant}\)
• C. \(\cot x + \text{constant}\)
• D. \(\cot^{-1} x + \text{constant}\)

Hint:

Remember the general form: \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\). Here \(a = 1\), directly giving \(\tan^{-1} x + C\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a fundamental standard integral of an algebraic rational function resulting directly from the inverse trigonometric derivative of the arctangent function.
Key Formula or Approach:
\[ \frac{d}{dx}\left(\tan^{-1} x\right) = \frac{1}{1 + x^2} \implies \int \frac{dx}{1 + x^2} = \tan^{-1} x + C \]

Step 2: Detailed Explanation:
By definition of antiderivatives, since the derivative of \(\tan^{-1} x\) with respect to \(x\) is \(\frac{1}{1 + x^2}\), integrating \(\frac{1}{1 + x^2}\) with respect to \(x\) yields \(\tan^{-1} x + C\), where \(C\) is the arbitrary constant of integration.

Step 3: Final Answer:
Hence, the indefinite integral evaluates to \(\tan^{-1} x + \text{constant}\), corresponding to option (B).