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

Answer

To solve the integral \( \int \frac{x^3 + 3}{x^3 - 3x} \, dx \), we will follow these steps:

### Step 1: Long Division
Since the degree of the numerator is the same as the degree of the denominator, we will perform polynomial long division.

1. Divide \( x^3 + 3 \) by \( x^3 - 3x \):
- The first term is \( 1 \) (since \( x^3 \div x^3 = 1 \)).
- Multiply \( 1 \) by \( x^3 - 3x \) to get \( x^3 - 3x \).
- Subtract this from \( x^3 + 3 \):
\[ (x^3 + 3) - (x^3 - 3x) = 3 + 3x \]

Thus, we can rewrite the integral as:
\[ \int \left( 1 + \frac{3 + 3x}{x^3 - 3x} \right) \, dx \]

### Step 2: Simplify the Integral
Now we have:
\[ \int 1 \, dx + \int \frac{3 + 3x}{x^3 - 3x} \, dx \]

### Step 3: Factor the Denominator
Factor \( x^3 - 3x \):
\[ x^3 - 3x = x(x^2 - 3) = x(x - \sqrt{3})(x + \sqrt{3}) \]

### Step 4: Partial Fraction Decomposition
We will decompose \( \frac{3 + 3x}{x(x - \sqrt{3})(x + \sqrt{3})} \):
\[ \frac{3 + 3x}{x(x - \sqrt{3})(x + \sqrt{3})} = \frac{A}{x} + \frac{B}{x - \sqrt{3}} + \frac{C}{x + \sqrt{3}} \]

Multiplying through by the denominator:
\[ 3 + 3x = A(x - \sqrt{3})(x + \sqrt{3}) + Bx(x + \sqrt{3}) + Cx(x - \sqrt{3}) \]

### Step 5: Solve for Coefficients
To find \( A, B, \) and \( C \), we can substitute convenient values for \( x \):

1. Let \( x = 0 \):
\[ 3 = A(-\sqrt{3})(\sqrt{3}) \implies A = -\frac{3}{3} = -1 \]

2. Let \( x = \sqrt{3} \):
\[ 3 + 3\sqrt{3} = B\sqrt{3}(\sqrt{3} + \sqrt{3}) \implies 3 + 3\sqrt{3} = 6B \implies B = \frac{1 + \sqrt{3}}{2} \]

3. Let \( x = -\sqrt{3} \):
\[ 3 - 3\sqrt{3} = C(-\sqrt{3})(-\sqrt{3} - \sqrt{3}) \implies 3 - 3\sqrt{3} = 6C \implies C = \frac{1 - \sqrt{3}}{2} \]

### Step 6: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left( 1 - \frac{1}{x} + \frac{(1 + \sqrt{3})/2}{x - \sqrt{3}} + \frac{(1 - \sqrt{3})/2}{x + \sqrt{3}} \right) \, dx \]

### Step 7: Integrate Each Term
Now we can integrate each term separately:
1. \( \int 1 \, dx = x \)
2. \( \int -\frac{1}{x} \, dx = -\ln |x| \)
3. \( \int \frac{(1 + \sqrt{3})/2}{x - \sqrt{3}} \, dx = \frac{(1 + \sqrt{3})}{2} \ln |x - \sqrt{3}| \)
4. \( \int \frac{(1 - \sqrt{3})/2}{x + \sqrt{3}} \, dx = \frac{(1 - \sqrt{3})}{2} \ln |x + \sqrt{3}| \)

### Step 8: Combine Results
Combining all the results, we have:
\[ \int \frac{x^3 + 3}{x^3 - 3x} \, dx = x - \ln |x| + \frac{(1 + \sqrt{3})}{2} \ln |x - \sqrt{3}| + \frac{(1 - \sqrt{3})}{2} \ln |x + \sqrt{3}| + C \]

### Final Answer
\[ \int \frac{x^3 + 3}{x^3 - 3x} \, dx = x - \ln |x| + \frac{(1 + \sqrt{3})}{2} \ln |x - \sqrt{3}| + \frac{(1 - \sqrt{3})}{2} \ln |x + \sqrt{3}| + C \]