Question:

<p><strong>Step 1:</strong> Rewrite the integral by extracting x³ from the radical:</p><p>$$I = \int \frac{dx}{x^2(x^4+1)^{3/4}} = \int \frac{dx}{x^2 \cdot x^3(1+\frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5(1+x^{-4})^{3/4}}$$</p><p><strong>Step 2:</strong> Divide numerator and denominator by x³:</p><p>$$I = \int \frac{x^{-3}dx}{x^2(x^4+1)^{3/4}} = \int \frac{x^{-3}}{(x^4+1)^{3/4}}dx$$</p><p><strong>Step 3:</strong> Use substitution u = x⁻⁴ + 1, so du = -4x⁻⁵dx, giving x⁻³dx = -¼x⁻¹du:</p><p>$$I = \int \frac{1}{u^{3/4}} \cdot \frac{-1}{4}du = -\frac{1}{4}\int u^{-3/4}du$$</p><p><strong>Step 4:</strong> Integrate using the power rule:</p><p>$$I = -\frac{1}{4} \cdot \frac{u^{1/4}}{1/4} + C = -u^{1/4} + C$$</p><p><strong>Step 5:</strong> Substitute back u = 1 + x⁻⁴:</p><p>$$I = -(1 + x^{-4})^{1/4} + C = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + C$$</p><p>$$\boxed{I = -\frac{(x^4+1)^{1/4}}{x} + C}$$</p>