Question

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

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

Hint:

In integrals involving square roots, always check whether the numerator resembles the derivative of the expression inside the root. That often reduces the integral directly to: \[ \int \frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)} \]

Answer 1

The Correct Option is C

Concept: Whenever the denominator contains a complicated algebraic expression under a square root, we try to manipulate the integrand so that the numerator becomes proportional to the derivative of the quantity inside the square root. This converts the integral into the standard form: \[ \int \frac{f'(x)}{\sqrt{f(x)}}dx \]

Step 1: Simplifying the square root expression.
Take \(x^4\) common from the expression inside the square root: \[ \sqrt{2x^4-2x^2+1} = x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} \] Substituting this into the integral: \[ I=\int \frac{x^2-1}{x^3\cdot x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}\,dx \] \[ I=\int \frac{x^2-1}{x^5\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}\,dx \] Divide numerator term-by-term: \[ I=\int \frac{\frac{1}{x^3}-\frac{1}{x^5}} {\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}\,dx \]

Step 2: Choosing a suitable substitution.
Let \[ t=2-\frac{2}{x^2}+\frac{1}{x^4} \] Differentiate both sides: \[ dt=\left(\frac{4}{x^3}-\frac{4}{x^5}\right)dx \] \[ dt=4\left(\frac{1}{x^3}-\frac{1}{x^5}\right)dx \] Therefore, \[ \left(\frac{1}{x^3}-\frac{1}{x^5}\right)dx=\frac{dt}{4} \]

Step 3: Substituting into the integral.
Now the integral becomes: \[ I=\int \frac{1}{\sqrt{t}}\cdot\frac{dt}{4} \] \[ I=\frac14\int t^{-1/2}dt \] Using the standard formula: \[ \int t^{-1/2}dt=2t^{1/2} \] we get: \[ I=\frac14(2\sqrt{t})+C \] \[ I=\frac12\sqrt{t}+C \] Substituting back: \[ I= \frac12\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C \] Hence, \[ \boxed{ \frac12\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C } \]