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 can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{x}{x + 1 + e^x} \, dx \]
We can manipulate the integrand by adding and subtracting \( (1 + e^x) \):
\[ I = \int \frac{x + 1 + e^x - (1 + e^x)}{x + 1 + e^x} \, dx \]
This simplifies to:
\[ I = \int \left( 1 - \frac{1 + e^x}{x + 1 + e^x} \right) \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two parts:
\[ I = \int 1 \, dx - \int \frac{1 + e^x}{x + 1 + e^x} \, dx \]
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 3: Simplify the Second Integral
Next, we focus on the second integral:
\[ \int \frac{1 + e^x}{x + 1 + e^x} \, dx \]
To evaluate this integral, we can use substitution. Let:
\[ t = x + 1 + e^x \]
Differentiating \( t \) gives:
\[ dt = (1 + e^x) \, dx \quad \Rightarrow \quad dx = \frac{dt}{1 + e^x} \]
Substituting into the integral, we have:
\[ \int \frac{1 + e^x}{t} \cdot \frac{dt}{1 + e^x} = \int \frac{1}{t} \, dt \]
This integral evaluates to:
\[ \ln |t| = \ln |x + 1 + e^x| \]

### Step 4: Combine Results
Now we can combine the results of both integrals:
\[ I = x - \ln |x + 1 + e^x| + C \]

### Step 5: Identify Coefficients
From the expression \( I = px + q \ln |x + 1 + e^x| + C \), we can identify:
- \( p = 1 \)
- \( q = -1 \)

### Step 6: Calculate \( p + q \)
Now we compute \( p + q \):
\[ p + q = 1 - 1 = 0 \]

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