Question

\( \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} dx = \)

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

Hint:

Try to find a function whose derivative matches the integrand. Often, the form of the answer choices can provide a hint for the substitution or differentiation approach. In this case, the term \( \frac{1}{x^2} \sqrt{2x^4 - 2x^2 + 1} \) suggests differentiating this form to see if it leads to the integrand.

Answer 1

The Correct Option is B

Let the integral be \( I = \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} dx \).
Divide the numerator and denominator by \( x^3 \): $$ I = \int \frac{\frac{x^2}{x^3} - \frac{1}{x^3}}{\sqrt{\frac{2x^4}{x^6} - \frac{2x^2}{x^6} + \frac{1}{x^6}}} dx = \int \frac{\frac{1}{x} - \frac{1}{x^3}}{\sqrt{\frac{2}{x^2} - \frac{2}{x^4} + \frac{1}{x^6}}} dx $$ Multiply the numerator and denominator by \( x^3 \): $$ I = \int \frac{x^2 - 1}{x^3 \cdot \frac{1}{|x|^3} \sqrt{2x^4 - 2x^2 + 1}} dx $$ This doesn't seem to simplify nicely.
Let's try a different approach.
Divide the numerator and denominator by \( x^2 \): $$ I = \int \frac{1 - \frac{1}{x^2}}{x \sqrt{2x^4 - 2x^2 + 1}} dx $$ Multiply the numerator and denominator by \( x \): $$ I = \int \frac{x - \frac{1}{x}}{\sqrt{x^2 (2x^4 - 2x^2 + 1)}} \frac{1}{x} dx = \int \frac{x - \frac{1}{x}}{\sqrt{2x^6 - 2x^4 + x^2}} \frac{1}{x} dx $$ Still not simplifying.
Let's go back to the original form and try a substitution related to the term under the square root.
Consider \( \frac{d}{dx} \left( \frac{1}{x^2} \sqrt{2x^4 - 2x^2 + 1} \right) \).
\( \frac{d}{dx} (x^{-2} (2x^4 - 2x^2 + 1)^{1/2}) \) \( = -2x^{-3} (2x^4 - 2x^2 + 1)^{1/2} + x^{-2} \cdot \frac{1}{2} (2x^4 - 2x^2 + 1)^{-1/2} (8x^3 - 4x) \) \( = \frac{-2 \sqrt{2x^4 - 2x^2 + 1}}{x^3} + \frac{(4x^3 - 2x)}{x^2 \sqrt{2x^4 - 2x^2 + 1}} \) \( = \frac{-2 (2x^4 - 2x^2 + 1) + x (4x^3 - 2x)}{x^3 \sqrt{2x^4 - 2x^2 + 1}} \) \( = \frac{-4x^4 + 4x^2 - 2 + 4x^4 - 2x^2}{x^3 \sqrt{2x^4 - 2x^2 + 1}} \) \( = \frac{2x^2 - 2}{x^3 \sqrt{2x^4 - 2x^2 + 1}} = 2 \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} \) So, \( \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} dx = \frac{1}{2} \frac{1}{x^2} \sqrt{2x^4 - 2x^2 + 1} + C = \frac{1}{2x^2} \sqrt{2x^4 - 2x^2 + 1} + C \).