What is `int(xe^(x)dx)/((x+1)^(2))` equal to ?
Where. C is the constant of integration.

Options

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

Answer

To solve the integral \( I = \int \frac{x e^x}{(x+1)^2} \, dx \), we can use integration by parts and some algebraic manipulation. Here’s a step-by-step solution:

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

### Step 2: Split the Fraction
We can rewrite the integrand by separating \( x \) in the numerator:
\[ I = \int \frac{(x+1 - 1)e^x}{(x+1)^2} \, dx = \int \left( \frac{(x+1)e^x}{(x+1)^2} - \frac{e^x}{(x+1)^2} \right) \, dx \]
This simplifies to:
\[ I = \int \left( \frac{e^x}{x+1} - \frac{e^x}{(x+1)^2} \right) \, dx \]

### Step 3: Break into Two Integrals
Now we can split the integral:
\[ I = \int \frac{e^x}{x+1} \, dx - \int \frac{e^x}{(x+1)^2} \, dx \]

### Step 4: Solve the First Integral
The first integral can be solved using the formula:
\[ \int e^x f(x) \, dx = e^x f(x) - \int e^x f'(x) \, dx \]
Here, let \( f(x) = \ln(x+1) \) and \( f'(x) = \frac{1}{x+1} \). Thus:
\[ \int \frac{e^x}{x+1} \, dx = e^x \ln(x+1) - \int e^x \cdot \frac{1}{x+1} \, dx \]

### Step 5: Solve the Second Integral
The second integral can also be solved using integration by parts:
\[ \int \frac{e^x}{(x+1)^2} \, dx \]
Using the same method as above, we can express it in terms of \( e^x \) and \( \ln(x+1) \).

### Step 6: Combine Results
Putting it all together, we have:
\[ I = e^x \ln(x+1) - \left( e^x \cdot \frac{1}{x+1} \right) + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = e^x \left( \ln(x+1) - \frac{1}{x+1} \right) + C \]