The value of `int((1+x))/(x(1+x e^(x))^(2))dx`,
is equal to

Options

A. (a) `log|(x)/(1+x e^(x))|+(1)/((1+x e^(x)))+C`
B. (b) `log|(x e^(x))/(1+x e^(x))|+(1)/(1+x e^(x))+C`
C. (c) `log |(x e^(x))/(1+e^(x))|+(1)/(1+x e^(x))+C`
D. (d) None of the above

Answer

To solve the integral \( I = \int \frac{1+x}{x(1+x e^x)^2} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1+x}{x(1+x e^x)^2} \, dx \]

### Step 2: Substitution
Let \( t = 1 + x e^x \). Then we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = e^x + x e^x = e^x (1 + x) \]
Thus, we have:
\[ dt = e^x (1+x) \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x (1+x)} \]

### Step 3: Express \( x \) in terms of \( t \)
From our substitution, we can express \( x \) in terms of \( t \):
\[ x e^x = t - 1 \quad \Rightarrow \quad x = \frac{t - 1}{e^x} \]

### Step 4: Substitute in the Integral
Substituting \( dx \) and \( t \) back into the integral, we need to express everything in terms of \( t \):
\[ I = \int \frac{1+x}{x(1+x e^x)^2} \cdot \frac{dt}{e^x (1+x)} \]
This simplifies to:
\[ I = \int \frac{dt}{(t-1)t^2} \]

### Step 5: Partial Fraction Decomposition
We can decompose \( \frac{1}{(t-1)t^2} \) using partial fractions:
\[ \frac{1}{(t-1)t^2} = \frac{A}{t-1} + \frac{B}{t} + \frac{C}{t^2} \]
Multiplying both sides by \( (t-1)t^2 \) and equating coefficients, we can solve for \( A \), \( B \), and \( C \).

### Step 6: Solve for Coefficients
1. Set \( t = 1 \):
\[ 1 = A(1^2) + B(1)(0) + C(0) \quad \Rightarrow \quad A = 1 \]
2. Set \( t = 0 \):
\[ 1 = A(0) + B(0) + C(-1) \quad \Rightarrow \quad C = -1 \]
3. Set \( t = -1 \):
\[ 1 = A(1) + B(-1) + C(1) \quad \Rightarrow \quad 1 = A - B - 1 \quad \Rightarrow \quad B = A - 2 \]

After solving, we find \( A = 1 \), \( B = 1 \), and \( C = -1 \). Thus:
\[ \frac{1}{(t-1)t^2} = \frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2} \]

### Step 7: Integrate Each Term
Now we can integrate each term:
\[ I = \int \left( \frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2} \right) dt \]
This gives us:
\[ I = \log |t-1| - \log |t| + \frac{1}{t} + C \]

### Step 8: Substitute Back
Substituting back \( t = 1 + x e^x \):
\[ I = \log |(1 + x e^x) - 1| - \log |1 + x e^x| + \frac{1}{1 + x e^x} + C \]
This simplifies to:
\[ I = \log |x e^x| - \log |1 + x e^x| + \frac{1}{1 + x e^x} + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = \log \left| \frac{x e^x}{1 + x e^x} \right| + \frac{1}{1 + x e^x} + C \]