Read the following text and answer the followig questions on the basis of the same :
`inte^(x)[f(x) + f'(x)]dx = int e^(x)f(x)dx + int e^(x) f'(x)dx`
`= f(x)e^(x) - int f'(x)e^(x)dx + int f'(x)e^(x)dx`
`= e^(x)f(x) + c`
`int(x e^(x))/((1+x)^(2))dx` = ________.

Options

A. `x e^(x) + c`
B. `(e^(x))/((x+1)^(2)) + c`
C. `(x e^(x))/(x+1) + c`
D. `(e^(x))/(x+1) + c`

Answer

To solve the integral \( \int \frac{x e^x}{(1+x)^2} \, dx \), we can use the technique of integration by parts and the given information in the problem statement.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ I = \int \frac{x e^x}{(1+x)^2} \, dx \]
We can split \( x \) as \( (1+x) - 1 \):
\[ I = \int \frac{(1+x)e^x - e^x}{(1+x)^2} \, dx \]
This simplifies to:
\[ I = \int \frac{(1+x)e^x}{(1+x)^2} \, dx - \int \frac{e^x}{(1+x)^2} \, dx \]

2. **Simplify the First Integral**:
The first integral simplifies as follows:
\[ I_1 = \int \frac{(1+x)e^x}{(1+x)^2} \, dx = \int \frac{e^x}{1+x} \, dx \]
Thus, we have:
\[ I = I_1 - I_2 \]
where \( I_2 = \int \frac{e^x}{(1+x)^2} \, dx \).

3. **Use Integration by Parts**:
For \( I_1 \), we can use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let \( u = \frac{1}{1+x} \) and \( dv = e^x \, dx \). Then, we have:
\[ du = -\frac{1}{(1+x)^2} \, dx \quad \text{and} \quad v = e^x \]
Therefore:
\[ I_1 = \left( \frac{e^x}{1+x} \right) - \int e^x \left(-\frac{1}{(1+x)^2}\right) \, dx \]
This gives:
\[ I_1 = \frac{e^x}{1+x} + I_2 \]

4. **Combine the Integrals**:
Substituting \( I_1 \) back into the equation for \( I \):
\[ I = \left( \frac{e^x}{1+x} + I_2 \right) - I_2 \]
This simplifies to:
\[ I = \frac{e^x}{1+x} \]

5. **Final Result**:
Therefore, the final result for the integral is:
\[ I = \frac{e^x}{1+x} + C \]