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Explanation

To solve the integral \( \int \frac{dx}{x(x^2 + 1)} \), we will use the method of partial fractions. Here’s a step-by-step solution:

### Step 1: Set up the partial fraction decomposition
We want to express \( \frac{1}{x(x^2 + 1)} \) in the form:
\[ \frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \]
where \( A \), \( B \), and \( C \) are constants to be determined.

**Hint:** Remember that the degree of the numerator must be less than the degree of the denominator in each term.

### Step 2: Combine the right-hand side
Multiply both sides by the common denominator \( x(x^2 + 1) \):
\[ 1 = A(x^2 + 1) + (Bx + C)x \]
This simplifies to:
\[ 1 = Ax^2 + A + Bx^2 + Cx \]
Combining like terms gives:
\[ 1 = (A + B)x^2 + Cx + A \]

**Hint:** Group the coefficients of like powers of \( x \) to set up equations.

### Step 3: Set up the system of equations
From the equation \( 1 = (A + B)x^2 + Cx + A \), we can equate coefficients:
1. For \( x^2 \): \( A + B = 0 \)
2. For \( x \): \( C = 0 \)
3. Constant term: \( A = 1 \)

**Hint:** Solve the equations systematically, starting from the constant term.

### Step 4: Solve for \( A \), \( B \), and \( C \)
From \( A = 1 \), we substitute into the first equation:
\[ 1 + B = 0 \implies B = -1 \]
And from the second equation, we already have \( C = 0 \).

Thus, we have:
- \( A = 1 \)
- \( B = -1 \)
- \( C = 0 \)

**Hint:** Make sure to substitute back to check if the values satisfy the original equation.

### Step 5: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{dx}{x(x^2 + 1)} = \int \left( \frac{1}{x} - \frac{x}{x^2 + 1} \right) dx \]

**Hint:** Break the integral into two separate integrals.

### Step 6: Integrate each term
Now we integrate each term separately:
1. \( \int \frac{1}{x} \, dx = \ln |x| \)
2. For \( \int \frac{x}{x^2 + 1} \, dx \), we can use the substitution \( u = x^2 + 1 \), \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \):
\[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln |x^2 + 1| \]

Thus, we have:
\[ \int \frac{dx}{x(x^2 + 1)} = \ln |x| - \frac{1}{2} \ln |x^2 + 1| + C \]

**Hint:** Remember to combine the logarithmic terms if needed.

### Final Answer
The final result is:
\[ \int \frac{dx}{x(x^2 + 1)} = \ln |x| - \frac{1}{2} \ln |x^2 + 1| + C \]