The given integral is:

$\int \frac{8x^3 - 2x}{\sqrt{9 - 4x^4}} \, dx$

Solution:

1. Substitution: Let $u = 9 - 4x^4$. Then: $du = -16x^3 \, dx$ Rearranging: $x^3 \, dx = -\frac{1}{16} \, du$

   Also, observe that $\sqrt{9 - 4x^4} = \sqrt{u}$.

2. Rewriting the integral: Substitute $u$ into the integral: $\int \frac{8x^3 - 2x}{\sqrt{9 - 4x^4}} \, dx = \int \frac{-8 \cdot \frac{1}{16} \, du - 2x}{\sqrt{u}}$

   Since $u = 9 - 4x^4$, isolate terms involving $x$ and simplify further. For detailed computation, I'll solve and provide a clean final form. sympy import symbols, integrate, sqrt

Define variables

x, u = symbols('x u')

Given integral

integrand = (8x**3 - 2x) / sqrt(9 - 4*x**4)

Perform substitution and solve

result = integrate(integrand, x) resultThe solution to the integral is:

$-\sqrt{9 - 4x^4} - 2 \cdot \text{Piecewise}\left( \left( -\frac{i}{4} \cosh^{-1}\left(\frac{2x^2}{3}\right), \frac{4|x^4|}{9} > 1 \right), \left( \frac{1}{4} \sin^{-1}\left(\frac{2x^2}{3}\right), \text{otherwise} \right) \right)$

This includes special cases for the behavior of the arcsine ($\sin^{-1}$) and arccosh ($\cosh^{-1}$) functions depending on the range of $x$. Let me know if you'd like a breakdown of how these functions contribute or need further clarification.

Follow-Up Questions:

1. How does substitution help simplify integrals like this one?
2. What are the conditions for switching between $\sin^{-1}$ and $\cosh^{-1}$ in such integrals?
3. Why does the square root function appear in the result?
4. What is the domain of the arcsine and arccosh functions in this solution?
5. How can this integral be applied in real-world scenarios?

Tip: Always consider domain restrictions when solving integrals involving square roots or inverse trigonometric functions to ensure proper interpretation of the result.