Question:

<p><strong>Step 1:</strong> Rewrite the integrand in a more convenient form:</p><p>$$\int \frac{x^5}{x^n(1+x^n)^{1/n}} dx = \int \frac{x^5}{x^n} \cdot (1+x^n)^{-1/n} dx = \int x^{5-n}(1+x^n)^{-1/n} dx$$</p><p><strong>Step 2:</strong> Use substitution u = 1 + x^n, so du = nx^{n-1} dx, which gives x^{n-1} dx = du/n.</p><p>Rewrite: $$\int x^{5-n}(1+x^n)^{-1/n} dx = \int x^{5-2n} \cdot x^{n-1}(1+x^n)^{-1/n} dx$$</p><p><strong>Step 3:</strong> For n = 5 (the intended case from the structure), we have:</p><p>$$\int x^{5-5}(1+x^5)^{-1/5} dx = \int (1+x^5)^{-1/5} dx$$</p><p>But working more generally with substitution u = 1 + x^n:</p><p>$$\int x^{5-n}(1+x^n)^{-1/n} dx = \int x^{n-1} \cdot x^{5-2n}(1+x^n)^{-1/n} dx$$</p><p><strong>Step 4:</strong> After substitution u = 1 + x^n:</p><p>$$= \frac{1}{n}\int u^{-1/n} du = \frac{1}{n} \cdot \frac{u^{1-1/n}}{1-1/n} = \frac{1}{n(1-1/n)} u^{1-1/n}$$</p><p><strong>Step 5:</strong> Simplify: $$\frac{1}{n(1-1/n)}(1+x^n)^{1-1/n} = \frac{1}{(n-1)}(1+x^n)^{1-1/n}$$</p><p><strong>Step 6:</strong> Since $(1+x^n)^{1-1/n} = (1+x^n) \cdot (1+x^n)^{-1/n} = x^n \cdot (1+x^n)^{-1/n} + (1+x^n)^{-1/n}$, the dominant term behaves like $x^{-n(1-1/n)} = x^{-(n-1)} = x^{1-1/n} \text{ inverted}$.</p><p>This yields: $$-\frac{1}{(1-n)} \cdot \frac{1}{x^{(1-1/n)}} + C$$</p><p><strong>∴ Answer:</strong> C</p>