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

Answer

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

### Step 1: Simplify the Denominator
Notice that the denominator can be factored:
\[ x^2 + 2x + 1 = (x + 1)^2 \]
Thus, we can rewrite the integral as:
\[ \int \frac{x}{(x + 1)^2} \, dx \]

### Step 2: Split the Numerator
Next, we can express \( x \) in terms of \( (x + 1) \):
\[ x = (x + 1) - 1 \]
Now, substitute this into the integral:
\[ \int \frac{(x + 1) - 1}{(x + 1)^2} \, dx = \int \left( \frac{x + 1}{(x + 1)^2} - \frac{1}{(x + 1)^2} \right) \, dx \]
This simplifies to:
\[ \int \left( \frac{1}{x + 1} - \frac{1}{(x + 1)^2} \right) \, dx \]

### Step 3: Integrate Each Term
Now we can integrate each term separately:
1. The integral of \( \frac{1}{x + 1} \) is:
\[ \int \frac{1}{x + 1} \, dx = \log |x + 1| + C_1 \]
2. The integral of \( -\frac{1}{(x + 1)^2} \) is:
\[ \int -\frac{1}{(x + 1)^2} \, dx = \frac{1}{x + 1} + C_2 \]

### Step 4: Combine the Results
Putting it all together, we have:
\[ \int \frac{x}{(x + 1)^2} \, dx = \log |x + 1| - \frac{1}{x + 1} + C \]
where \( C = C_1 + C_2 \) is a constant of integration.

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{x}{x^2 + 2x + 1} \, dx = \log |x + 1| - \frac{1}{x + 1} + C \]

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