Question

If \(\int \frac{dx}{x^2(x^n + 1)^{(n-1)/n}} = -[f(x)]^{1/n} + c\) then \(f(x)\) is

• A. \(1 + x^n\)
• B. \(1 + x^{-n}\)
• C. \(x^n + x^{-n}\)
• D. None of these

Hint:

Factor out \(x^n\) from the denominator to simplify.

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ \int \frac{dx}{x^2(x^n + 1)^{(n-1)/n}} = \int \frac{dx}{x^{n+1}(1 + x^{-n})^{(n-1)/n}} \]
Step 2: Calculation / Simplification}
Let \(t = 1 + x^{-n} \Rightarrow dt = -n x^{-n-1} dx \Rightarrow \frac{dx}{x^{n+1}} = -\frac{dt}{n}\)
\(\int \frac{-dt/n}{t^{(n-1)/n}} = -\frac{1}{n} \int t^{-(n-1)/n} dt = -\frac{1}{n} \cdot \frac{t^{1/n}}{1/n} + c = -t^{1/n} + c\)
\(= -(1 + x^{-n})^{1/n} + c\)
\(\therefore f(x) = 1 + x^{-n}\)
Step 3: Final Answer
\[ 1 + x^{-n} \]