Evaluate : `int (x)/(x+1)^(2)(dx)`

Answer

To evaluate the integral \( \int \frac{x}{(x+1)^2} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral:**
\[ \int \frac{x}{(x+1)^2} \, dx \]
We can manipulate the numerator by adding and subtracting 1:
\[ \int \frac{x + 1 - 1}{(x+1)^2} \, dx = \int \frac{(x + 1) - 1}{(x+1)^2} \, dx \]

2. **Separate the Integral:**
\[ \int \frac{x + 1}{(x+1)^2} \, dx - \int \frac{1}{(x+1)^2} \, dx \]
This simplifies to:
\[ \int \frac{1}{x+1} \, dx - \int \frac{1}{(x+1)^2} \, dx \]

3. **Evaluate the First Integral:**
The integral \( \int \frac{1}{x+1} \, dx \) is:
\[ \ln |x + 1| \]

4. **Evaluate the Second Integral:**
The integral \( \int \frac{1}{(x+1)^2} \, dx \) can be computed using the power rule:
\[ -\frac{1}{x + 1} \]

5. **Combine the Results:**
Now, substituting back, we have:
\[ \ln |x + 1| + \frac{1}{x + 1} + C \]
where \( C \) is the constant of integration.

### Final Answer:
\[ \int \frac{x}{(x+1)^2} \, dx = \ln |x + 1| + \frac{1}{x + 1} + C \]