To evaluate the integral \( \int \frac{3x + 1}{2x^2 - 2x + 3} \, dx \), we can follow these steps:
### Step 1: Rewrite the Integral
We start with the integral:
\[
I = \int \frac{3x + 1}{2x^2 - 2x + 3} \, dx
\]
### Step 2: Perform Polynomial Long Division (if necessary)
In this case, the degree of the numerator (1) is less than the degree of the denominator (2), so we can proceed directly without long division.
### Step 3: Use Partial Fraction Decomposition
Since the denominator \( 2x^2 - 2x + 3 \) does not factor nicely (its discriminant is negative), we can express the numerator in terms of the derivative of the denominator:
\[
3x + 1 = A(4x - 2) + B
\]
where \( A \) and \( B \) are constants to be determined.
### Step 4: Differentiate the Denominator
The derivative of the denominator \( 2x^2 - 2x + 3 \) is:
\[
\frac{d}{dx}(2x^2 - 2x + 3) = 4x - 2
\]
### Step 5: Set Up the Equation
Now, we can equate coefficients:
\[
3x + 1 = A(4x - 2) + B
\]
Expanding this gives:
\[
3x + 1 = 4Ax - 2A + B
\]
### Step 6: Compare Coefficients
From the equation \( 3x + 1 = (4A)x + (-2A + B) \), we can set up the following system:
1. \( 4A = 3 \)
2. \( -2A + B = 1 \)
### Step 7: Solve for A and B
From the first equation, we find:
\[
A = \frac{3}{4}
\]
Substituting \( A \) into the second equation:
\[
-2\left(\frac{3}{4}\right) + B = 1 \implies -\frac{3}{2} + B = 1 \implies B = 1 + \frac{3}{2} = \frac{5}{2}
\]
### Step 8: Substitute Back into the Integral
Now we can rewrite the integral:
\[
I = \int \left( \frac{3}{4} \cdot \frac{4x - 2}{2x^2 - 2x + 3} + \frac{5}{2} \cdot \frac{1}{2x^2 - 2x + 3} \right) \, dx
\]
This can be split into two separate integrals:
\[
I = \frac{3}{4} \int \frac{4x - 2}{2x^2 - 2x + 3} \, dx + \frac{5}{2} \int \frac{1}{2x^2 - 2x + 3} \, dx
\]
### Step 9: Solve the First Integral
For the first integral, we can use substitution:
Let \( t = 2x^2 - 2x + 3 \), then \( dt = (4x - 2) \, dx \).
Thus, we have:
\[
\frac{3}{4} \int \frac{dt}{t} = \frac{3}{4} \ln |t| + C_1 = \frac{3}{4} \ln |2x^2 - 2x + 3| + C_1
\]
### Step 10: Solve the Second Integral
For the second integral, we complete the square in the denominator:
\[
2x^2 - 2x + 3 = 2\left(x^2 - x + \frac{3}{2}\right) = 2\left((x - \frac{1}{2})^2 + \frac{5}{4}\right)
\]
Thus, we have:
\[
\int \frac{1}{2\left((x - \frac{1}{2})^2 + \frac{5}{4}\right)} \, dx = \frac{1}{\sqrt{5}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C_2
\]
### Step 11: Combine Results
Putting it all together, we have:
\[
I = \frac{3}{4} \ln |2x^2 - 2x + 3| + \frac{5}{2\sqrt{5}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C
\]
### Final Answer
Thus, the final answer is:
\[
\int \frac{3x + 1}{2x^2 - 2x + 3} \, dx = \frac{3}{4} \ln |2x^2 - 2x + 3| + \frac{5}{2\sqrt{5}} \tan^{-1}\left(\frac{2x - 1}{\sqrt{5}}\right) + C
\]
