If `int(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) `x^n-x^-n`

Options

A. `(1+x^(n))`
B. `1+x^(-n) `
C. `x^(n)+x^(-n) `
D. non of these

Explanation

To solve the given problem, we need to evaluate the integral

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

and find the function \( f(x) \).

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \frac{dx}{x^2 (x^n + 1)^{(n-1)/n}} \]

### Step 2: Substitute
Let’s make a substitution to simplify the integral. We can set:

\[ t = x^n + 1 \implies dt = n x^{n-1} dx \implies dx = \frac{dt}{n x^{n-1}} \]

From our substitution, we can express \( x^{n-1} \) in terms of \( t \):

\[ x^n = t - 1 \implies x = (t - 1)^{1/n} \]

### Step 3: Change the Limits
Now we need to express \( x^2 \) in terms of \( t \):

\[ x^2 = ((t - 1)^{1/n})^2 = (t - 1)^{2/n} \]

Now substituting \( dx \) and \( x^2 \) into the integral:

\[ I = \int \frac{1}{(t - 1)^{2/n} t^{(n-1)/n}} \cdot \frac{dt}{n (t - 1)^{(n-1)/n}} \]

### Step 4: Simplify the Integral
This simplifies to:

\[ I = \frac{1}{n} \int \frac{dt}{(t - 1)^{(n+1)/n} t^{(n-1)/n}} \]

### Step 5: Solve the Integral
Now we can integrate:

\[ I = -\frac{1}{n} \left( t^{1/n} \right) + C \]

### Step 6: Substitute Back
Substituting back \( t = x^n + 1 \):

\[ I = -\frac{1}{n} \left( (x^n + 1)^{1/n} \right) + C \]

### Step 7: Identify \( f(x) \)
From the equation given in the problem, we have:

\[ -(f(x))^{1/n} = -\frac{1}{n} (x^n + 1)^{1/n} + C \]

Thus, we can identify:

\[ f(x) = 1 + x^n \]

### Final Answer
The function \( f(x) \) is:

\[ \boxed{1 + x^n} \]