Evaluate
`int (x^(3))/(x+2)dx`

Answer

To evaluate the integral \( \int \frac{x^3}{x+2} \, dx \), we can use substitution and polynomial long division. Here’s a step-by-step solution:

### Step 1: Polynomial Long Division
First, we perform polynomial long division on \( \frac{x^3}{x+2} \).

1. Divide the leading term \( x^3 \) by the leading term \( x \) of the divisor \( x + 2 \):
\[ x^3 \div x = x^2 \]
2. Multiply \( x^2 \) by \( x + 2 \):
\[ x^2(x + 2) = x^3 + 2x^2 \]
3. Subtract this from \( x^3 \):
\[ x^3 - (x^3 + 2x^2) = -2x^2 \]
4. Now bring down the next term (which is 0 since we have no \( x \) term yet):
\[ -2x^2 \]
5. Divide \( -2x^2 \) by \( x \):
\[ -2x^2 \div x = -2x \]
6. Multiply \( -2x \) by \( x + 2 \):
\[ -2x(x + 2) = -2x^2 - 4x \]
7. Subtract this from \( -2x^2 \):
\[ -2x^2 - (-2x^2 - 4x) = 4x \]
8. Now bring down the next term (which is 0):
\[ 4x \]
9. Divide \( 4x \) by \( x \):
\[ 4x \div x = 4 \]
10. Multiply \( 4 \) by \( x + 2 \):
\[ 4(x + 2) = 4x + 8 \]
11. Subtract this from \( 4x \):
\[ 4x - (4x + 8) = -8 \]

So, we can express the integral as:
\[ \frac{x^3}{x+2} = x^2 - 2x + 4 - \frac{8}{x+2} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{x^3}{x+2} \, dx = \int \left( x^2 - 2x + 4 - \frac{8}{x+2} \right) \, dx \]

### Step 3: Integrate Each Term
Now we can integrate term by term:
1. \( \int x^2 \, dx = \frac{x^3}{3} \)
2. \( \int -2x \, dx = -x^2 \)
3. \( \int 4 \, dx = 4x \)
4. \( \int -\frac{8}{x+2} \, dx = -8 \ln |x+2| \)

Putting it all together, we have:
\[ \int \frac{x^3}{x+2} \, dx = \frac{x^3}{3} - x^2 + 4x - 8 \ln |x+2| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{x^3}{x+2} \, dx = \frac{x^3}{3} - x^2 + 4x - 8 \ln |x+2| + C \]