Question

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

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

Hint:

Look for an expression whose derivative matches the integrand — trying derivatives of expressions involving square roots and rational terms can lead you quickly to the answer.

Answer 1

The Correct Option is B

We are given: \[ \int \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx \] Let us simplify the numerator: \[ x^4 - 1 = (x^2)^2 - 1 = (x^2 - 1)(x^2 + 1) \] Now let: \[ I = \int \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx = \int \frac{(x^2 - 1)(x^2 + 1)}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx \] Let us try substitution: \[ u = x^2 \Rightarrow du = 2x \, dx \Rightarrow dx = \frac{du}{2x} \] But notice: \[ x^4 + x^2 + 1 = (x^2)^2 + x^2 + 1 = u^2 + u + 1 \] Instead of substitution, try inspection: Let us test the derivative of \(\frac{\sqrt{x^4 + x^2 + 1}}{x}\): 
Let \(f(x) = \frac{\sqrt{x^4 + x^2 + 1}}{x}\) 
Use quotient rule: \[ f'(x) = \frac{x \cdot \frac{1}{2\sqrt{x^4 + x^2 + 1}}(4x^3 + 2x) - \sqrt{x^4 + x^2 + 1}}{x^2} \] Multiply and simplify confirms the integrand: \[ \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} = \frac{d}{dx} \left( \frac{\sqrt{x^4 + x^2 + 1}}{x} \right) \] Therefore, \[ \int \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx = \boxed{\frac{\sqrt{x^4 + x^2 + 1}}{x} + c} \]