If $$\int \frac{dx}{x^2(x^n + 1)^{(n-1)/n}} = -[f(x)]^{1/n} + C$$, then $f(x)$ is
Step-by-Step Solution
Key Concept: Recognize that the integrand can be rewritten by factoring out x^(-n) to expose a substitution pattern. Use substitution u = 1 + x^(-n) to transform the integral into a standard power function form ∫u^(-(n-1)/n)du, which yields u^(1/n).
Rewrite the integral as $\int x^{-2}(x^n + 1)^{-(n-1)/n}dx = \int x^{-2} \cdot x^{-n}(1 + x^{-n})^{-(n-1)/n}dx$. Substitute $u = 1 + x^{-n}$, so $du = -nx^{-n-1}dx$. The integral becomes $\frac{1}{n}\int u^{-(n-1)/n}du = \frac{1}{n} \cdot \frac{u^{1/n}}{1/n} + C = \frac{1}{1/n - 1}u^{1/n} + C = \frac{u^{1/n}}{(1-n)/n} + C$.
Correct Answer: 2