`inte^(x)*(x)/((1+x)^(2))dx=?`

Options

A. `e^(x)*(1)/((1+x))+C`
B. `e^(x)*(1)/(x)+C`
C. `e^(x)*(x)/((1+x))+C`
D. none of these

Explanation

To evaluate the integral \( \int \frac{e^x \cdot x}{(1+x)^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \frac{e^x \cdot x}{(1+x)^2} \, dx \]
We can express \( x \) as \( (1+x) - 1 \):
\[ I = \int \frac{e^x \cdot ((1+x) - 1)}{(1+x)^2} \, dx \]
This simplifies to:
\[ I = \int \frac{e^x \cdot (1+x)}{(1+x)^2} \, dx - \int \frac{e^x}{(1+x)^2} \, dx \]
Thus, we can separate the integral into two parts:
\[ I = \int \frac{e^x}{1+x} \, dx - \int \frac{e^x}{(1+x)^2} \, dx \]

### Step 2: Define the First Integral
Let’s denote the first integral as \( I_1 \):
\[ I_1 = \int \frac{e^x}{1+x} \, dx \]

### Step 3: Use Integration by Parts for \( I_1 \)
To solve \( I_1 \), we will use integration by parts. We choose:
- \( u = \frac{1}{1+x} \) (first function)
- \( dv = e^x \, dx \) (second function)

Then, we compute \( du \) and \( v \):
\[ du = -\frac{1}{(1+x)^2} \, dx \quad \text{and} \quad v = e^x \]

Using integration by parts:
\[ I_1 = uv - \int v \, du \]
Substituting the values:
\[ I_1 = \frac{e^x}{1+x} - \int e^x \left(-\frac{1}{(1+x)^2}\right) \, dx \]
This simplifies to:
\[ I_1 = \frac{e^x}{1+x} + \int \frac{e^x}{(1+x)^2} \, dx \]

### Step 4: Substitute \( I_1 \) Back into \( I \)
Now, substituting \( I_1 \) back into the expression for \( I \):
\[ I = \left( \frac{e^x}{1+x} + \int \frac{e^x}{(1+x)^2} \, dx \right) - \int \frac{e^x}{(1+x)^2} \, dx \]
The integrals \( \int \frac{e^x}{(1+x)^2} \, dx \) cancel out:
\[ I = \frac{e^x}{1+x} + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{e^x \cdot x}{(1+x)^2} \, dx = \frac{e^x}{1+x} + C \]