What is `intlog(x+1)` dx is equal to ?

Options

A. A) `x log (x+1)-x+c`
B. B) `(x+1)log(x+1)-x+c`
C. C) `(1)/(x+1)+c`
D. D) `(log(x+1))/(x+1)+c`

Answer

To solve the integral \( \int \log(x + 1) \, dx \), we will use the method of integration by parts. Here’s a step-by-step solution:

### Step 1: Set up the integral
Let
\[ I = \int \log(x + 1) \, dx \]

### Step 2: Choose \( u \) and \( dv \)
We will use integration by parts, where we choose:
- \( u = \log(x + 1) \)
- \( dv = dx \)

### Step 3: Differentiate \( u \) and integrate \( dv \)
Now, we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{1}{x + 1} \, dx \)
- \( v = x \)

### Step 4: Apply the integration by parts formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we substitute our values:
\[ I = x \log(x + 1) - \int x \cdot \frac{1}{x + 1} \, dx \]

### Step 5: Simplify the integral
Now we simplify the integral:
\[ \int x \cdot \frac{1}{x + 1} \, dx = \int \left( 1 - \frac{1}{x + 1} \right) \, dx \]
This can be split into two separate integrals:
\[ \int 1 \, dx - \int \frac{1}{x + 1} \, dx \]

### Step 6: Compute the integrals
Calculating these integrals:
1. \( \int 1 \, dx = x \)
2. \( \int \frac{1}{x + 1} \, dx = \log(x + 1) \)

Thus,
\[ \int x \cdot \frac{1}{x + 1} \, dx = x - \log(x + 1) \]

### Step 7: Substitute back into the equation
Substituting back into our equation for \( I \):
\[ I = x \log(x + 1) - (x - \log(x + 1)) \]
This simplifies to:
\[ I = x \log(x + 1) - x + \log(x + 1) \]

### Step 8: Combine like terms
Combining the terms gives:
\[ I = (x + 1) \log(x + 1) - x + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final result is:
\[ \int \log(x + 1) \, dx = (x + 1) \log(x + 1) - x + C \]

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