`int(dx)/(x(x^n+1))` is equal to

Options

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

Explanation

To solve the integral \( \int \frac{dx}{x(x^n + 1)} \), we will follow these steps:

### Step 1: Substitution
Let \( t = x^n \). Then, differentiating both sides gives us:
\[ dt = n x^{n-1} dx \quad \Rightarrow \quad dx = \frac{dt}{n x^{n-1}}. \]
Since \( x = t^{1/n} \), we have:
\[ x^{n-1} = (t^{1/n})^{n-1} = t^{(n-1)/n}. \]
Thus, we can express \( dx \) in terms of \( t \):
\[ dx = \frac{dt}{n t^{(n-1)/n}} = \frac{dt}{n t^{(n-1)/n}}. \]

### Step 2: Rewrite the Integral
Substituting \( dx \) into the integral gives us:
\[ \int \frac{dx}{x(x^n + 1)} = \int \frac{1}{x(t + 1)} \cdot \frac{dt}{n t^{(n-1)/n}}. \]
Since \( x = t^{1/n} \), we have:
\[ \int \frac{1}{t^{1/n}(t + 1)} \cdot \frac{dt}{n t^{(n-1)/n}} = \frac{1}{n} \int \frac{dt}{t(t + 1)}. \]

### Step 3: Partial Fraction Decomposition
We can decompose \( \frac{1}{t(t + 1)} \) using partial fractions:
\[ \frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1}. \]
Multiplying through by \( t(t + 1) \) gives:
\[ 1 = A(t + 1) + Bt. \]
Setting \( t = 0 \) gives \( A = 1 \). Setting \( t = -1 \) gives \( B = -1 \). Thus, we have:
\[ \frac{1}{t(t + 1)} = \frac{1}{t} - \frac{1}{t + 1}. \]

### Step 4: Integrate
Now we can integrate:
\[ \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t + 1} \right) dt = \frac{1}{n} \left( \log |t| - \log |t + 1| \right) + C. \]
This simplifies to:
\[ \frac{1}{n} \log \left| \frac{t}{t + 1} \right| + C. \]

### Step 5: Back Substitute
Recalling that \( t = x^n \), we substitute back:
\[ \frac{1}{n} \log \left| \frac{x^n}{x^n + 1} \right| + C. \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{dx}{x(x^n + 1)} = \frac{1}{n} \log \left| \frac{x^n}{x^n + 1} \right| + C. \]