Question 40

Mathematics Definite Integrals Hard

The value of \(\int \frac{({x}^2-1)}{{x}^3\sqrt[]{2{x}^4-2{x}^2+1}}dx\)

(A) \(2\sqrt[]{2-\frac{2}{{x}^2}+\frac{1}{{x}^4}}+C\)
(B) \(2\sqrt[]{2+\frac{2}{{x}^2}+\frac{1}{{x}^4}}+C\)
(C) \(\frac{1}{2}\sqrt[]{2-\frac{2}{{x}^2}+\frac{1}{{x}^4}}+C\)
(D) None of the above
✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

The integration can be written as: \[ \int \frac{\left(x^2 - 1\right) dx}{x^5 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} = \int \frac{\left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}}. \] Now, let: \[ 2 - \frac{2}{x^2} + \frac{1}{x^4} = t \quad \Rightarrow \quad \left(\frac{4}{x^3} - \frac{4}{x^5}\right) dx = dt. \] Simplifying: \[ \left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx = \frac{dt}{4}. \] The integration becomes: \[ \int \frac{1}{4} \frac{dt}{\sqrt{t}} = \frac{1}{2} \sqrt{t} + C. \] Substitute back for \( t \): \[ \frac{1}{2} \sqrt{t} + C = \frac{1}{2} \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + C. \]

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