Evaluate: `int(x^3)/(x+1)dx`

Explanation

To evaluate the integral

\[ \int \frac{x^3}{x+1} \, dx, \]

we will follow these steps:

### Step 1: Perform Long Division
Since the degree of the numerator \(x^3\) is greater than the degree of the denominator \(x+1\), we will perform polynomial long division.

1. Divide \(x^3\) by \(x\) to get \(x^2\).
2. Multiply \(x^2\) by \(x + 1\) to get \(x^3 + x^2\).
3. Subtract this from \(x^3\):

\[ x^3 - (x^3 + x^2) = -x^2. \]

4. Bring down the next term, which is 0 (since there is no \(x\) term in the original numerator), resulting in \(-x^2\).
5. Divide \(-x^2\) by \(x\) to get \(-x\).
6. Multiply \(-x\) by \(x + 1\) to get \(-x^2 - x\).
7. Subtract this from \(-x^2\):

\[ -x^2 - (-x^2 - x) = x. \]

8. Bring down the next term, which is 0 (since there is no constant term in the original numerator), resulting in \(x\).
9. Divide \(x\) by \(x\) to get \(1\).
10. Multiply \(1\) by \(x + 1\) to get \(x + 1\).
11. Subtract this from \(x\):

\[ x - (x + 1) = -1. \]

Thus, we can express the original fraction as:

\[ \frac{x^3}{x+1} = x^2 - x + 1 - \frac{1}{x+1}. \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral:

\[ \int \frac{x^3}{x+1} \, dx = \int \left( x^2 - x + 1 - \frac{1}{x+1} \right) \, dx. \]

### Step 3: Integrate Each Term
Now we will integrate each term separately:

1. \(\int x^2 \, dx = \frac{x^3}{3}\),
2. \(\int -x \, dx = -\frac{x^2}{2}\),
3. \(\int 1 \, dx = x\),
4. \(\int -\frac{1}{x+1} \, dx = -\ln |x+1|\).

Putting it all together, we have:

\[ \int \frac{x^3}{x+1} \, dx = \frac{x^3}{3} - \frac{x^2}{2} + x - \ln |x+1| + C, \]

where \(C\) is the constant of integration.

### Final Answer
Thus, the final result is:

\[ \int \frac{x^3}{x+1} \, dx = \frac{x^3}{3} - \frac{x^2}{2} + x - \ln |x+1| + C. \]

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