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

Answer

To solve the integral \(\int \frac{dx}{x^3 - 1}\), we will use the method of partial fractions. Here is a step-by-step solution:

### Step 1: Factor the Denominator
The first step is to factor the denominator \(x^3 - 1\). We can use the difference of cubes formula:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]
Thus, we can rewrite the integral as:
\[ \int \frac{dx}{(x - 1)(x^2 + x + 1)} \]

### Step 2: Set Up Partial Fractions
Next, 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 to be determined.

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

### Step 4: Expand the Right Side
Expanding the right side gives:
\[ 1 = A(x^2 + x + 1) + Bx^2 - Bx + Cx - C \]
Combining like terms results in:
\[ 1 = (A + B)x^2 + (A - B + C)x + (A - C) \]

### Step 5: Set Up a System of Equations
Now we equate coefficients from both sides:
1. \(A + B = 0\) (coefficient of \(x^2\))
2. \(A - B + C = 0\) (coefficient of \(x\))
3. \(A - C = 1\) (constant term)

### Step 6: Solve the System of Equations
From the first equation, we have:
\[ B = -A \]
Substituting \(B\) into the second equation:
\[ A - (-A) + C = 0 \implies 2A + C = 0 \implies C = -2A \]
Substituting \(C\) into the third equation:
\[ A - (-2A) = 1 \implies 3A = 1 \implies A = \frac{1}{3} \]
Then,
\[ B = -A = -\frac{1}{3}, \quad C = -2A = -\frac{2}{3} \]

### Step 7: Rewrite the Integral
Now we can rewrite the integral using the values of \(A\), \(B\), and \(C\):
\[ \int \left(\frac{1/3}{x - 1} + \frac{-\frac{1}{3}x - \frac{2}{3}}{x^2 + x + 1}\right) dx \]
This can be separated into two integrals:
\[ \frac{1}{3} \int \frac{dx}{x - 1} - \frac{1}{3} \int \frac{x + 2}{x^2 + x + 1} dx \]

### Step 8: Solve the Integrals
1. The first integral:
\[ \frac{1}{3} \ln |x - 1| \]

2. For the second integral, we can split it:
\[ -\frac{1}{3} \int \frac{x}{x^2 + x + 1} dx - \frac{2}{3} \int \frac{dx}{x^2 + x + 1} \]
- The first part can be solved using substitution \(u = x^2 + x + 1\).
- The second part can be solved by completing the square in the denominator.

Completing the square:
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]
Thus, the integral becomes:
\[ -\frac{1}{3} \ln |x^2 + x + 1| - \frac{2}{3} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) \]

### Final Answer
Putting it all together, we have:
\[ \int \frac{dx}{x^3 - 1} = \frac{1}{3} \ln |x - 1| - \frac{1}{3} \ln |x^2 + x + 1| - \frac{2}{3\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C \]