Evaluate the following Integrals :
`int (dx)/(x(x^(n)+1))`

Answer

To evaluate the integral

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

we can follow these steps:

### Step 1: Substitution
Let \( t = x^{n+1} \). Then, we differentiate to find \( dx \):
\[ dt = (n x^n) dx \quad \Rightarrow \quad dx = \frac{dt}{n x^n}. \]

### Step 2: Express \( x \) in terms of \( t \)
From our substitution \( t = x^{n+1} \), we can express \( x \) as:
\[ x = t^{\frac{1}{n+1}}. \]
Thus, \( x^n = (t^{\frac{1}{n+1}})^n = t^{\frac{n}{n+1}} \).

### Step 3: Substitute into the integral
Now, substituting \( x \) and \( dx \) into the integral:
\[ I = \int \frac{1}{t^{\frac{1}{n+1}} \left(t + 1\right)} \cdot \frac{dt}{n t^{\frac{n}{n+1}}}. \]
This simplifies to:
\[ I = \frac{1}{n} \int \frac{dt}{t^{\frac{n+1}{n+1}}(t + 1)} = \frac{1}{n} \int \frac{dt}{t(t + 1)}. \]

### Step 4: Partial Fraction Decomposition
Now, we can use 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 5: Integrate
Now we can integrate:
\[ I = \frac{1}{n} \left( \int \frac{1}{t} dt - \int \frac{1}{t + 1} dt \right). \]
This results in:
\[ I = \frac{1}{n} \left( \ln |t| - \ln |t + 1| \right) + C. \]

### Step 6: Substitute back for \( t \)
Substituting back \( t = x^{n+1} \):
\[ I = \frac{1}{n} \left( \ln |x^{n+1}| - \ln |x^{n+1} + 1| \right) + C. \]
Using properties of logarithms, we can combine this:
\[ I = \frac{1}{n} \ln \left| \frac{x^{n+1}}{x^{n+1} + 1} \right| + C. \]

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