Question

∫ (x2 - 1)dx/(x3(2x4 - 2x2 1)1/2) = ?

Answer 1

1. Simplify the Integrand:
∫ (x² - 1) dx / (x³ √(2x⁴ - 2x² + 1))

2. Factor out x⁴ from under the square root:
∫ (x² - 1) dx / (x³ √(x⁴(2 - 2/x² + 1/x⁴)))
∫ (x² - 1) dx / (x³ * x² √(2 - 2/x² + 1/x⁴))
∫ (x² - 1) dx / (x⁵ √(2 - 2/x² + 1/x⁴))

3. Divide numerator and denominator by x⁴:
∫ (1/x² - 1/x⁴) dx / (x √(2 - 2/x² + 1/x⁴))

4. Make a Substitution:
Let u = 2 - 2/x² + 1/x⁴
Then du = (4/x³ - 4/x⁵) dx = 4(1/x³ - 1/x⁵) dx

Notice that (1/x² - 1/x⁴) dx = (1/4) * (4/x³ - 4/x⁵) * x dx = (1/4) du * x

Also, from u = 2 - 2/x² + 1/x⁴, multiply both sides by x^4, giving u x^4 = 2x^4 - 2x^2 + 1.

Then, x = sqrt( (2x^4 - 2x^2 + 1)/u ) = sqrt ( (2- 2/x^2 + 1/x^4 ) x^4 / u ) = sqrt (ux^4/u)

We need to rewrite (1/x^2 - 1/x^4) dx = 1/4 (4/x^3 - 4/x^5) dx.
Multiply numerator and denominator by x.

∫ (x(1/x^2 - 1/x^4)) dx / (x^2 √(2 - 2/x² + 1/x⁴))
∫ (1/x - 1/x^3) dx / (x^2 √(2 - 2/x² + 1/x⁴))

Let v = 1/x^2. dv = -2/x^3 dx. (1/x - 1/x^3) = (1/x - dv/(-2))

Then let's go back and use u = 2 - 2/x² + 1/x⁴.

∫ (1/x² - 1/x⁴) dx / (x √(2 - 2/x² + 1/x⁴)) = ∫ (x² - 1)/(x^5 sqrt(2 - 2/x^2 + 1/x^4) ) dx

Rewrite the numerator as:

(x^2 - 1) / x^5 = 1/x^3 - 1/x^5

∫ (1/x^3 - 1/x^5) dx / √(2 - 2/x² + 1/x⁴)

Let u = 2 - 2/x² + 1/x⁴

du = (4/x³ - 4/x⁵) dx

(1/4) du = (1/x³ - 1/x⁵) dx

∫ (1/4) du / √u = (1/4) ∫ u^(-1/2) du = (1/4) * 2 * u^(1/2) + C = (1/2) √u + C

(1/2) √(2 - 2/x² + 1/x⁴) + C = (1/2) √((2x⁴ - 2x² + 1) / x⁴) + C

= (1/2x²) √(2x⁴ - 2x² + 1) + C

5. Final Answer:
(1/2x²)√(2x⁴ - 2x² + 1) + C