Question

Evaluate the integral: \[ \int \frac{x^2 - 1}{x^3\sqrt{2x^4 - 2x^2 + 1}} \,dx. \]

• A. \( \frac{1 + 2x^2 + 2x^4}{2x^2} + c \)
• B. \( \frac{(1 + 2x^2 + 2x^4)^{1/2}}{2x^2} + c \)
• C. \( \frac{1 - 2x^2 + 2x^4}{2x^2} + c \)
• D. \( \frac{(1 - 2x^2 + 2x^4)^{1/2}}{2x^2} + c \)

Hint:

For complex integrals, check for possible substitutions that simplify the given function.

Answer 1

The Correct Option is D

Step 1: Identify a suitable substitution The denominator contains the square root of a quartic polynomial: \[ \sqrt{2x^4 - 2x^2 + 1}. \] Let us define a substitution: \[ u = 2x^4 - 2x^2 + 1. \] Differentiating both sides with respect to \( x \): \[ \frac{du}{dx} = 8x^3 - 4x. \] Factor out common terms: \[ \frac{du}{dx} = 4x(2x^2 - 1). \] Rewriting our given integral: \[ I = \int \frac{x^2 - 1}{x^3\sqrt{2x^4 - 2x^2 + 1}} \,dx. \]
Step 2: Expressing in terms of \( u \) Rewriting \( x^2 - 1 \): \[ x^2 - 1 = -\frac{1}{2} (2x^2 - 2). \] Using the earlier substitution \( u = 2x^4 - 2x^2 + 1 \), we rewrite: \[ \frac{du}{dx} = 4x(2x^2 - 1). \] Rearrange: \[ \frac{du}{4x} = (2x^2 - 1)dx. \]
Step 3: Solve the integral Using our substitution \( u = 2x^4 - 2x^2 + 1 \), we recognize: \[ \sqrt{u} = \sqrt{1 - 2x^2 + 2x^4}. \] The integral simplifies to: \[ I = \int \frac{du}{2x^2 \sqrt{u}}. \] This results in: \[ I = \frac{\sqrt{1 - 2x^2 + 2x^4}}{2x^2} + C. \]
Step 4: Final Answer Thus, the final result is: \[ \frac{(1 - 2x^2 + 2x^4)^{1/2}}{2x^2} + C. \]