`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:
\[ x^2 + 2x + 1 = (x + 1)^2 \]
So we rewrite the integral:
\[ \int \frac{x}{(x + 1)^2} \, dx \]

### Step 2: Use Partial Fraction Decomposition
We can express \( \frac{x}{(x + 1)^2} \) using partial fractions:
\[ \frac{x}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} \]
Multiplying through by the denominator \( (x + 1)^2 \) gives:
\[ x = A(x + 1) + B \]
Expanding the right side:
\[ x = Ax + A + B \]
Rearranging gives:
\[ x = Ax + (A + B) \]
From this, we can equate coefficients:
1. \( A = 1 \)
2. \( A + B = 0 \)

### Step 3: Solve for A and B
From \( A = 1 \), we substitute into the second equation:
\[ 1 + B = 0 \implies B = -1 \]
Thus, we can rewrite the integral as:
\[ \int \left( \frac{1}{x + 1} - \frac{1}{(x + 1)^2} \right) \, dx \]

### Step 4: Integrate Each Term
Now we can integrate each term separately:
1. The integral of \( \frac{1}{x + 1} \) is \( \ln |x + 1| \).
2. The integral of \( -\frac{1}{(x + 1)^2} \) is \( \frac{1}{x + 1} \).

Putting it all together, we have:
\[ \int \frac{x}{(x + 1)^2} \, dx = \ln |x + 1| + \frac{1}{x + 1} + C \]

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