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

Answer

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

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

### Step 2: Express the Numerator in Terms of the Derivative of the Denominator
We notice that the numerator \( 2x - 1 \) can be expressed as a linear combination of the derivative of the denominator \( 2x^2 + 2x + 1 \). The derivative is:
\[ \frac{d}{dx}(2x^2 + 2x + 1) = 4x + 2 \]
We can express \( 2x - 1 \) in terms of \( 4x + 2 \):
\[ 2x - 1 = A(4x + 2) + B \]
where \( A \) and \( B \) are constants we need to determine.

### Step 3: Determine Constants A and B
To find \( A \) and \( B \), we can compare coefficients:
1. The coefficient of \( x \): \( 4A = 2 \) implies \( A = \frac{1}{2} \).
2. The constant term: \( 2A + B = -1 \). Substituting \( A = \frac{1}{2} \):
\[ 2 \cdot \frac{1}{2} + B = -1 \implies 1 + B = -1 \implies B = -2 \]

### Step 4: Substitute A and B Back into the Integral
Now we can rewrite the integral:
\[ I = \int \frac{\frac{1}{2}(4x + 2) - 2}{2x^2 + 2x + 1} \, dx \]
This can be split into two separate integrals:
\[ I = \frac{1}{2} \int \frac{4x + 2}{2x^2 + 2x + 1} \, dx - 2 \int \frac{1}{2x^2 + 2x + 1} \, dx \]

### Step 5: Evaluate the First Integral
For the first integral, we can use the formula:
\[ \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C \]
where \( f(x) = 2x^2 + 2x + 1 \):
\[ \int \frac{4x + 2}{2x^2 + 2x + 1} \, dx = \ln |2x^2 + 2x + 1| + C \]
Thus,
\[ \frac{1}{2} \int \frac{4x + 2}{2x^2 + 2x + 1} \, dx = \frac{1}{2} \ln |2x^2 + 2x + 1| \]

### Step 6: Evaluate the Second Integral
For the second integral, we can complete the square in the denominator:
\[ 2x^2 + 2x + 1 = 2\left(x^2 + x + \frac{1}{2}\right) = 2\left((x + \frac{1}{2})^2 + \frac{1}{4}\right) \]
Thus,
\[ \int \frac{1}{2x^2 + 2x + 1} \, dx = \frac{1}{2} \int \frac{1}{(x + \frac{1}{2})^2 + \left(\frac{1}{2}\right)^2} \, dx \]
Using the formula:
\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \]
we have:
\[ = \frac{1}{2} \cdot \frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{2(x + \frac{1}{2})}{1}\right) = \tan^{-1}(2x + 1) \]

### Step 7: Combine the Results
Putting it all together, we have:
\[ I = \frac{1}{2} \ln |2x^2 + 2x + 1| - 2 \tan^{-1}(2x + 1) + C \]

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