If `int(x)/(x+1+e^(x))dx=px+qln|x+1+e^(x)|+c`, where c is the constant of integration, then `p+q` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the integral \( \int \frac{x}{x + 1 + e^x} \, dx \) and express it in the form \( px + q \ln|x + 1 + e^x| + c \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{x}{x + 1 + e^x} \, dx \]

### Step 2: Split the Numerator
We can rewrite the numerator \( x \) as:
\[ x = (x + 1 + e^x) - (1 + e^x) \]
Thus, we can express the integral as:
\[ I = \int \frac{(x + 1 + e^x) - (1 + e^x)}{x + 1 + e^x} \, dx \]
This simplifies to:
\[ I = \int 1 \, dx - \int \frac{1 + e^x}{x + 1 + e^x} \, dx \]

### Step 3: Integrate the First Part
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Focus on the Second Integral
Now we need to evaluate:
\[ \int \frac{1 + e^x}{x + 1 + e^x} \, dx \]
We can use substitution. Let:
\[ u = x + 1 + e^x \quad \Rightarrow \quad du = (1 + e^x) \, dx \]
Thus, we have:
\[ dx = \frac{du}{1 + e^x} \]
Substituting this into the integral gives:
\[ \int \frac{1 + e^x}{u} \cdot \frac{du}{1 + e^x} = \int \frac{1}{u} \, du = \ln|u| + C = \ln|x + 1 + e^x| + C \]

### Step 5: Combine the Results
Putting everything together, we have:
\[ I = x - \ln|x + 1 + e^x| + C \]

### Step 6: Compare with Given Form
Now, we can compare this with the given form:
\[ px + q \ln|x + 1 + e^x| + c \]
From our result:
\[ I = 1 \cdot x + (-1) \cdot \ln|x + 1 + e^x| + C \]
This gives us:
- \( p = 1 \)
- \( q = -1 \)

### Step 7: Calculate \( p + q \)
Now we find:
\[ p + q = 1 - 1 = 0 \]

### Final Answer
Thus, the value of \( p + q \) is:
\[ \boxed{0} \]