`intx^3/(x+1)dx`is equal to:

Options

A. `x+(x^(2))/(2)+(x^(3))/(3)-log|1-x|+C`
B. `x+(x^(2))/(2)-(x^(3))/(3)-log|1-x|+C`
C. `x-(x^(2))/(2)-(x^(3))/(3)-log|1+x|+C`
D. `x-(x^(2))/(2) + (x^(3))/(3) - log|1+x|+C`

Explanation

To solve the integral \( \int \frac{x^3}{x+1} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integrand:
\[ \int \frac{x^3}{x+1} \, dx = \int \left( x^3 + 1 - 1 \right) \frac{1}{x+1} \, dx = \int \left( \frac{x^3 + 1}{x+1} - \frac{1}{x+1} \right) \, dx \]

### Step 2: Separate the Integral
Now we can separate the integral into two parts:
\[ \int \frac{x^3 + 1}{x+1} \, dx - \int \frac{1}{x+1} \, dx \]
Let’s denote the first integral as \( I_1 \) and the second integral as \( I_2 \):
\[ I = I_1 - I_2 \]

### Step 3: Solve \( I_2 \)
The integral \( I_2 \) is straightforward:
\[ I_2 = \int \frac{1}{x+1} \, dx = \log |x+1| + C_1 \]

### Step 4: Solve \( I_1 \)
Next, we need to simplify \( I_1 \):
\[ I_1 = \int \frac{x^3 + 1}{x+1} \, dx \]
We can factor \( x^3 + 1 \) as \( (x+1)(x^2 - x + 1) \):
\[ I_1 = \int (x^2 - x + 1) \, dx \]

### Step 5: Integrate \( I_1 \)
Now we can integrate \( I_1 \):
\[ I_1 = \int (x^2 - x + 1) \, dx = \frac{x^3}{3} - \frac{x^2}{2} + x + C_2 \]

### Step 6: Combine Results
Now we can combine the results of \( I_1 \) and \( I_2 \):
\[ I = I_1 - I_2 = \left( \frac{x^3}{3} - \frac{x^2}{2} + x + C_2 \right) - \left( \log |x+1| + C_1 \right) \]
This simplifies to:
\[ I = \frac{x^3}{3} - \frac{x^2}{2} + x - \log |x+1| + C \]
where \( C = C_2 - C_1 \).

### Final Answer
Thus, the final result is:
\[ \int \frac{x^3}{x+1} \, dx = \frac{x^3}{3} - \frac{x^2}{2} + x - \log |x+1| + C \]

---