Verify the following using the concept of integration as an antiderivative.
`int x^3/(x+1) dx= x-x^2/2+x^3/3 -log|x+1|+C`

Answer

To verify the integral \(\int \frac{x^3}{x+1} \, dx = x - \frac{x^2}{2} + \frac{x^3}{3} - \log|x+1| + C\), we will differentiate the right-hand side and check if we obtain the left-hand side.

### Step 1: Differentiate the Right-Hand Side
We start by differentiating the expression on the right-hand side:
\[ y = x - \frac{x^2}{2} + \frac{x^3}{3} - \log|x+1| + C \]
Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(x\right) - \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}\left(\frac{x^3}{3}\right) - \frac{d}{dx}\left(\log|x+1|\right) \]

### Step 2: Apply the Derivative Rules
Using basic differentiation rules:
1. The derivative of \(x\) is \(1\).
2. The derivative of \(-\frac{x^2}{2}\) is \(-x\).
3. The derivative of \(\frac{x^3}{3}\) is \(x^2\).
4. The derivative of \(-\log|x+1|\) is \(-\frac{1}{x+1}\) (using the chain rule).

Putting it all together:
\[ \frac{dy}{dx} = 1 - x + x^2 - \frac{1}{x+1} \]

### Step 3: Simplify the Expression
Now, we simplify the expression:
\[ \frac{dy}{dx} = 1 - x + x^2 - \frac{1}{x+1} \]
To combine the terms, we can express \(1\) as \(\frac{x+1}{x+1}\):
\[ \frac{dy}{dx} = \frac{x+1}{x+1} - x + x^2 - \frac{1}{x+1} \]
Combining the fractions:
\[ \frac{dy}{dx} = \frac{x + 1 - x(x + 1) + x^2(x + 1) - 1}{x + 1} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{x^3 + x^2 - x^2 - x - 1 + 1}{x + 1} = \frac{x^3}{x + 1} \]

### Step 4: Conclusion
Thus, we have:
\[ \frac{dy}{dx} = \frac{x^3}{x + 1} \]
This confirms that:
\[ \int \frac{x^3}{x + 1} \, dx = x - \frac{x^2}{2} + \frac{x^3}{3} - \log|x + 1| + C \]
is indeed correct.