Question

Evaluate: $\int \frac{1}{1 + x^2} \, dx$

• A. \( \sin^{-1}x + C \)
• B. \( \tan^{-1}x + C \)
• C. \( \log(1 + x^2) + C \)
• D. \( \sec^{-1}x + C \)

Hint:

Be careful not to confuse this with the logarithmic form! An integral like \(\int \frac{2x}{1 + x^2} \, dx\) evaluates to \(\log(1+x^2) + C\) via substitution, because the derivative of the denominator is present in the numerator. However, without that \(x\) variable on top, the form is strictly an inverse trigonometric arctan curve.

Answer 1

The Correct Option is B

Concept: This is a core standard integral form derived directly from basic differential calculus rules. Integration is the inverse process of differentiation. If the derivative of a function \(F(x)\) is \(f(x)\): \[ \frac{d}{dx}[F(x)] = f(x) \quad \implies \quad \int f(x) \, dx = F(x) + C \]

Step 1: Connecting with standard trigonometric derivatives.
Recall the standard derivative formula for inverse trigonometric functions, specifically the inverse tangent function: \[ \frac{d}{dx}(\tan^{-1}x) = \frac{1}{1 + x^2} \]

Step 2: Applying the indefinite integral mapping.
By reversing this derivative relationship, the anti-derivative is found immediately: \[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}x + C \]