Evaluate: `int (x^(2))/(x^(2)-4) dx`

Answer

To evaluate the integral \( I = \int \frac{x^2}{x^2 - 4} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We can rewrite the integrand by adding and subtracting 4 in the numerator:
\[ I = \int \frac{x^2 - 4 + 4}{x^2 - 4} \, dx \]
This can be separated into two parts:
\[ I = \int \frac{x^2 - 4}{x^2 - 4} \, dx + \int \frac{4}{x^2 - 4} \, dx \]

### Step 2: Simplify the first integral
The first part simplifies to:
\[ \int \frac{x^2 - 4}{x^2 - 4} \, dx = \int 1 \, dx = x \]

### Step 3: Simplify the second integral
Now, we need to evaluate the second integral:
\[ \int \frac{4}{x^2 - 4} \, dx \]
We can factor the denominator:
\[ x^2 - 4 = (x - 2)(x + 2) \]
Thus, we can rewrite the integral as:
\[ 4 \int \frac{1}{(x - 2)(x + 2)} \, dx \]

### Step 4: Use partial fraction decomposition
We can express \( \frac{1}{(x - 2)(x + 2)} \) using partial fractions:
\[ \frac{1}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2} \]
Multiplying through by the denominator \( (x - 2)(x + 2) \) gives:
\[ 1 = A(x + 2) + B(x - 2) \]
Setting \( x = 2 \):
\[ 1 = A(4) \implies A = \frac{1}{4} \]
Setting \( x = -2 \):
\[ 1 = B(-4) \implies B = -\frac{1}{4} \]
Thus, we have:
\[ \frac{1}{(x - 2)(x + 2)} = \frac{1/4}{x - 2} - \frac{1/4}{x + 2} \]

### Step 5: Integrate the partial fractions
Now we can integrate:
\[ 4 \int \left( \frac{1/4}{x - 2} - \frac{1/4}{x + 2} \right) \, dx = \int \left( \frac{1}{x - 2} - \frac{1}{x + 2} \right) \, dx \]
This gives:
\[ \int \frac{1}{x - 2} \, dx - \int \frac{1}{x + 2} \, dx = \ln |x - 2| - \ln |x + 2| = \ln \left| \frac{x - 2}{x + 2} \right| \]

### Step 6: Combine results
Combining all parts, we have:
\[ I = x + 4 \left( \ln \left| \frac{x - 2}{x + 2} \right| \right) + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = x + 4 \ln \left| \frac{x - 2}{x + 2} \right| + C \]