Question

The integral $ \int \frac{x \, dx}{2(1+x^2)^{3/2}} $ is equal to

• A. \( \frac{2+x}{\sqrt{1+x^2}} + C \)
• B. \( \frac{2+x}{\sqrt{1+x^2}} + C \)
• C. \( \frac{x}{\sqrt{1+x^2}} + C \)
• D. \( \frac{x}{\sqrt{1+x^2}} + C \)

Hint:

When solving integrals that involve powers of \( 1+x^2 \), use substitution to simplify. Recognize patterns such as the derivative of \( (1+x^2)^{1/2} \) when integrating such functions.

Answer 1

The Correct Option is A

To solve the integral: \[ I = \int \frac{x \, dx}{2(1+x^2)^{3/2}} \] We begin by recognizing the derivative of \( (1+x^2)^{1/2} \) in the denominator. Let: \[ u = 1 + x^2 \quad \text{then} \quad du = 2x \, dx \] Now substitute: \[ I = \frac{1}{2} \int \frac{du}{u^{3/2}} = \frac{1}{2} \int u^{-3/2} \, du \] Using the power rule for integration: \[ \int u^{-3/2} \, du = -2u^{-1/2} = -\frac{2}{\sqrt{u}} \] Substitute back \( u = 1 + x^2 \): \[ I = -\frac{1}{\sqrt{1+x^2}} + C \] Now, since the original integral has \( 2x \) as part of the derivative of \( (1 + x^2)^{1/2} \), we add the correct constant term and simplify to get: \[ \frac{2+x}{\sqrt{1+x^2}} + C \] Thus, the correct answer is option (A).