`int(3x+1)/(2x^(2)+x-1)dx`

Answer

To solve the integral \( \int \frac{3x + 1}{2x^2 + x - 1} \, dx \), we will use the method of partial fractions and substitution. Let's go through the steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \( 2x^2 + x - 1 \).

To factor \( 2x^2 + x - 1 \), we look for two numbers that multiply to \( 2 \times -1 = -2 \) and add to \( 1 \). The numbers \( 2 \) and \( -1 \) work.

Thus, we can rewrite the quadratic as:
\[ 2x^2 + 2x - x - 1 = (2x^2 + 2x) + (-x - 1) = 2x(x + 1) - 1(x + 1) = (2x - 1)(x + 1) \]
So, we have:
\[ 2x^2 + x - 1 = (2x - 1)(x + 1) \]

### Step 2: Set Up Partial Fraction Decomposition
Next, we express \( \frac{3x + 1}{(2x - 1)(x + 1)} \) as a sum of partial fractions:
\[ \frac{3x + 1}{(2x - 1)(x + 1)} = \frac{A}{2x - 1} + \frac{B}{x + 1} \]
Multiplying through by the denominator \( (2x - 1)(x + 1) \) gives:
\[ 3x + 1 = A(x + 1) + B(2x - 1) \]

### Step 3: Solve for A and B
Expanding the right side:
\[ 3x + 1 = Ax + A + 2Bx - B = (A + 2B)x + (A - B) \]
Now, we equate coefficients:
1. \( A + 2B = 3 \) (coefficient of \( x \))
2. \( A - B = 1 \) (constant term)

From the second equation, we can express \( A \) in terms of \( B \):
\[ A = B + 1 \]
Substituting into the first equation:
\[ (B + 1) + 2B = 3 \implies 3B + 1 = 3 \implies 3B = 2 \implies B = \frac{2}{3} \]
Now substituting \( B \) back to find \( A \):
\[ A = \frac{2}{3} + 1 = \frac{5}{3} \]

### Step 4: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{3x + 1}{(2x - 1)(x + 1)} \, dx = \int \left( \frac{5/3}{2x - 1} + \frac{2/3}{x + 1} \right) \, dx \]
This can be separated into two integrals:
\[ = \frac{5}{3} \int \frac{1}{2x - 1} \, dx + \frac{2}{3} \int \frac{1}{x + 1} \, dx \]

### Step 5: Integrate Each Term
Now we integrate each term:
1. For \( \int \frac{1}{2x - 1} \, dx \):
\[ = \frac{1}{2} \ln |2x - 1| + C_1 \]
2. For \( \int \frac{1}{x + 1} \, dx \):
\[ = \ln |x + 1| + C_2 \]

Combining these results:
\[ = \frac{5}{3} \cdot \frac{1}{2} \ln |2x - 1| + \frac{2}{3} \ln |x + 1| + C \]
\[ = \frac{5}{6} \ln |2x - 1| + \frac{2}{3} \ln |x + 1| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{3x + 1}{2x^2 + x - 1} \, dx = \frac{5}{6} \ln |2x - 1| + \frac{2}{3} \ln |x + 1| + C \]