`int (dx)/((x^(3)+1))`

Answer

To solve the integral \( \int \frac{dx}{x^3 + 1} \), we will use the method of partial fractions. Let's go through the solution step by step.

### Step 1: Factor the denominator
The expression \( x^3 + 1 \) can be factored using the sum of cubes formula:
\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \]
So, we can rewrite the integral as:
\[ \int \frac{dx}{(x + 1)(x^2 - x + 1)} \]

### Step 2: Set up partial fractions
We express the integrand as a sum of partial fractions:
\[ \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} \]
where \( A \), \( B \), and \( C \) are constants that we need to determine.

### Step 3: Clear the denominators
Multiply through by the denominator \((x + 1)(x^2 - x + 1)\):
\[ 1 = A(x^2 - x + 1) + (Bx + C)(x + 1) \]

### Step 4: Expand and collect like terms
Expanding the right-hand side:
\[ 1 = A(x^2 - x + 1) + Bx^2 + Bx + Cx + C \]
\[ = (A + B)x^2 + (-A + B + C)x + (A + C) \]

### Step 5: Set up equations for coefficients
Now we equate coefficients from both sides:
1. For \( x^2 \): \( A + B = 0 \)
2. For \( x \): \( -A + B + C = 0 \)
3. For the constant term: \( A + C = 1 \)

### Step 6: Solve the system of equations
From \( A + B = 0 \), we can express \( B \) as:
\[ B = -A \]

Substituting \( B \) into the second equation:
\[ -A - A + C = 0 \implies C = 2A \]

Now substituting \( C \) into the third equation:
\[ A + 2A = 1 \implies 3A = 1 \implies A = \frac{1}{3} \]

Using \( A \) to find \( B \) and \( C \):
\[ B = -A = -\frac{1}{3}, \quad C = 2A = \frac{2}{3} \]

### Step 7: Write the partial fractions
Now we can write the partial fractions:
\[ \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{1/3}{x + 1} + \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1} \]

### Step 8: Integrate each term
Now we can integrate:
\[ \int \left( \frac{1/3}{x + 1} + \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1} \right) dx \]

1. The first integral:
\[ \int \frac{1/3}{x + 1} \, dx = \frac{1}{3} \ln |x + 1| \]

2. The second integral can be split:
\[ \int \frac{-\frac{1}{3}x}{x^2 - x + 1} \, dx + \int \frac{\frac{2}{3}}{x^2 - x + 1} \, dx \]

For the first part, use substitution \( u = x^2 - x + 1 \) (with \( du = (2x - 1)dx \)). For the second part, complete the square in the denominator and use the arctangent formula.

### Step 9: Final result
After performing the integrations, we combine the results:
\[ \int \frac{dx}{x^3 + 1} = \frac{1}{3} \ln |x + 1| - \frac{1}{6} \ln |x^2 - x + 1| + \frac{2}{3} \tan^{-1} \left( \frac{x - \frac{1}{2}}{\sqrt{\frac{3}{4}}} \right) + C \]