The integral ∫ (x^2 - 1) / (x^3 * sqrt(2x^4 - 2x^2 + 1)) dx is equal to -
(A) (sqrt(2x^4 - 2x^2 + 1)) / x^2 + c
(B) (sqrt(2x^4 - 2x^2 + 1)) / x^3 + c
(C) (sqrt(2x^4 - 2x^2 + 1)) / x + c
(D) (sqrt(2x^4 - 2x^2 + 1)) / 2x^2 + c
Step-by-Step Solution
Key Concept: Divide numerator and denominator by x^3 to simplify the integral into a form suitable for substitution.
Step 1: Manipulate the integrand for simplification.
The given integral is $I = \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} dx$.
To simplify the expression under the square root, factor out $x^4$:
$$x^3 \sqrt{2x^4 - 2x^2 + 1} = x^3 \sqrt{x^4 \left(2 - \frac{2}{x^2} + \frac{1}{x^4}\right)}$$
Take $x^2$ out of the square root:
$$x^3 \cdot x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} = x^5 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}$$
Substitute this simplified denominator back into the integral:
$$I = \int \frac{x^2 - 1}{x^5 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx$$
Now, divide the numerator by $x^5$ to facilitate the upcoming substitution:
$$I = \int \frac{\frac{x^2 - 1}{x^5}}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{\frac{1}{x^3} - \frac{1}{x^5}}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx$$
Step 2: Apply a suitable substitution.
Let $t$ be the expression inside the square root in the denominator:
$$t = 2 - \frac{2}{x^2} + \frac{1}{x^4}$$
Now, find the differential $dt$ by differentiating $t$ with respect to $x$:
$$dt = \frac{d}{dx}\left(2 - 2x^{-2} + x^{-4}\right) dx$$
$$dt = \left(0 - 2(-2x^{-3}) + (-4)x^{-5}\right) dx$$
$$dt = \left(\frac{4}{x^3} - \frac{4}{x^5}\right) dx$$
Factor out 4 from the expression for $dt$:
$$dt = 4\left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx$$
From this, we can express the numerator term $\left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx$ in terms of $dt$:
$$\left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx = \frac{1}{4} dt$$
Step 3: Rewrite the integral in terms of $t$.
Substitute $t$ and $\left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx = \frac{1}{4} dt$ into the transformed integral from Step 1:
$$I = \int \frac{\frac{1}{4} dt}{\sqrt{t}}$$
$$I = \frac{1}{4} \int t^{-1/2} dt$$
Step 4: Integrate the expression in terms of $t$.
Perform the integration using the power rule for integrals, $\int u^n du = \frac{u^{n+1}}{n+1} + C$:
$$I = \frac{1}{4} \left(\frac{t^{-1/2 + 1}}{-1/2 + 1}\right) + C$$
$$I = \frac{1}{4} \left(\frac{t^{1/2}}{1/2}\right) + C$$
$$I = \frac{1}{4} (2\sqrt{t}) + C$$
$$I = \frac{1}{2}\sqrt{t} + C$$
Step 5: Substitute back to express the result in terms of $x$ and state the final answer.
Replace $t$ with its original expression in terms of $x$, $t = 2 - \frac{2}{x^2} + \frac{1}{x^4}$:
$$I = \frac{1}{2}\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + C$$
To match the format of the given options, combine the terms inside the square root by finding a common denominator:
$$I = \frac{1}{2}\sqrt{\frac{2x^4 - 2x^2 + 1}{x^4}} + C$$
Now, take $\sqrt{x^4} = x^2$ out of the square root in the denominator:
$$I = \frac{1}{2} \frac{\sqrt{2x^4 - 2x^2 + 1}}{x^2} + C$$
This result matches Option 4.
The final answer is $\boxed{\text{Option 4}}$
Correct Answer: 1