Evaluate: `int1/(x\ (x^4+1))\ dx`

Answer

To evaluate the integral \( \int \frac{1}{x(x^4 + 1)} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{x(x^4 + 1)} \, dx \]

### Step 2: Factor Out the Highest Degree Term
We can factor out \( x^4 \) from the denominator:
\[ I = \int \frac{1}{x \left( x^4 + 1 \right)} \, dx = \int \frac{1}{x^5 \left( 1 + \frac{1}{x^4} \right)} \, dx \]

### Step 3: Substitution
Now, we will use a substitution. Let:
\[ t = 1 + \frac{1}{x^4} \]
Then, we differentiate \( t \):
\[ dt = -\frac{4}{x^5} \, dx \quad \Rightarrow \quad dx = -\frac{x^5}{4} \, dt \]

### Step 4: Express \( dx \) in Terms of \( t \)
From our substitution, we can express \( \frac{dx}{x^5} \):
\[ \frac{dx}{x^5} = -\frac{1}{4} dt \]

### Step 5: Substitute Back into the Integral
Now we substitute \( t \) and \( dx \) into the integral:
\[ I = \int \frac{-1}{4} \frac{1}{t} \, dt \]

### Step 6: Integrate
The integral of \( \frac{1}{t} \) is:
\[ I = -\frac{1}{4} \ln |t| + C \]

### Step 7: Substitute Back for \( t \)
Now we substitute back \( t = 1 + \frac{1}{x^4} \):
\[ I = -\frac{1}{4} \ln \left| 1 + \frac{1}{x^4} \right| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{x(x^4 + 1)} \, dx = -\frac{1}{4} \ln \left| 1 + \frac{1}{x^4} \right| + C \]