To solve the integral \( \int \frac{x^3}{x^2 - 4} \, dx \), we can use polynomial long division followed by integration. Here’s a step-by-step solution:
### Step 1: Polynomial Long Division
We start by dividing \( x^3 \) by \( x^2 - 4 \).
1. Divide the leading term: \( \frac{x^3}{x^2} = x \).
2. Multiply \( x \) by \( x^2 - 4 \): \( x(x^2 - 4) = x^3 - 4x \).
3. Subtract this from \( x^3 \):
\[
x^3 - (x^3 - 4x) = 4x
\]
So, we can rewrite the integral as:
\[
\int \frac{x^3}{x^2 - 4} \, dx = \int \left( x + \frac{4x}{x^2 - 4} \right) \, dx
\]
### Step 2: Split the Integral
Now we can split the integral into two parts:
\[
\int \left( x + \frac{4x}{x^2 - 4} \right) \, dx = \int x \, dx + \int \frac{4x}{x^2 - 4} \, dx
\]
### Step 3: Integrate the First Part
The first integral is straightforward:
\[
\int x \, dx = \frac{x^2}{2}
\]
### Step 4: Integrate the Second Part
For the second integral, we can use substitution. Let \( t = x^2 - 4 \). Then, the derivative \( dt = 2x \, dx \) or \( x \, dx = \frac{1}{2} dt \).
Substituting into the integral gives:
\[
\int \frac{4x}{x^2 - 4} \, dx = 4 \int \frac{x}{t} \cdot \frac{1}{2} dt = 2 \int \frac{1}{t} \, dt
\]
### Step 5: Integrate the Logarithm
Now we can integrate:
\[
2 \int \frac{1}{t} \, dt = 2 \ln |t| + C = 2 \ln |x^2 - 4| + C
\]
### Step 6: Combine the Results
Putting it all together, we have:
\[
\int \frac{x^3}{x^2 - 4} \, dx = \frac{x^2}{2} + 2 \ln |x^2 - 4| + C
\]
### Final Answer
Thus, the final answer is:
\[
\int \frac{x^3}{x^2 - 4} \, dx = \frac{x^2}{2} + 2 \ln |x^2 - 4| + C
\]
