Question

What is the value of \(\int \frac{1}{x^2 + 1} \, dx\)?

• A. \(\ln|x + 1| + C\)
• B. \(\tan^{-1}x + C\)
• C. \(\ln|x - 1| + C\)
• D. \(\tan^{-1}(x^2) + C\)

Hint:

The integral \(\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\) is a key formula in calculus.

Answer 1

The Correct Option is B

The integral \(\int \frac{1}{x^2 + 1} \, dx\) is a standard form. Recognize that \(\frac{1}{x^2 + a^2}\) integrates to \(\frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\). Here, \(a = 1\), so: \[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}x + C \] Alternatively, use substitution: let \(x = \tan \theta\), then \(dx = \sec^2 \theta \, d\theta\), and the integral becomes: \[ \int \frac{\sec^2 \theta}{\tan^2 \theta + 1} \, d\theta = \int \frac{\sec^2 \theta}{\sec^2 \theta} \, d\theta = \int 1 \, d\theta = \theta + C = \tan^{-1}x + C \] Thus, option (2) is correct.