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

Options

A. `(1)/(5)tan^(-1)((4)/(3))`
B. `(1)/(5)log((4)/(3))`
C. `5 tan^(-1)((3)/(4))`
D. `5 log ((3)/(4))`

Answer

To solve the definite integral \( \int_{2}^{3} \frac{1}{2x^2 + 3x - 2} \, dx \), we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the quadratic expression in the denominator:

\[ 2x^2 + 3x - 2 \]

We can rearrange \( 3x \) as \( 4x - x \):

\[ 2x^2 + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2) \]

Thus, we can rewrite the integral as:

\[ \int_{2}^{3} \frac{1}{(2x - 1)(x + 2)} \, dx \]

### Step 2: Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on the integrand:

\[ \frac{1}{(2x - 1)(x + 2)} = \frac{A}{2x - 1} + \frac{B}{x + 2} \]

Multiplying through by the denominator \( (2x - 1)(x + 2) \):

\[ 1 = A(x + 2) + B(2x - 1) \]

Expanding and rearranging gives:

\[ 1 = Ax + 2A + 2Bx - B = (A + 2B)x + (2A - B) \]

Setting coefficients equal, we have the system:

1. \( A + 2B = 0 \)
2. \( 2A - B = 1 \)

From the first equation, \( A = -2B \). Substituting into the second equation:

\[ 2(-2B) - B = 1 \implies -4B - B = 1 \implies -5B = 1 \implies B = -\frac{1}{5} \]

Substituting back to find \( A \):

\[ A = -2(-\frac{1}{5}) = \frac{2}{5} \]

Thus, we can rewrite the integral as:

\[ \int_{2}^{3} \left( \frac{2/5}{2x - 1} - \frac{1/5}{x + 2} \right) \, dx \]

### Step 3: Integrate Each Term
Now we can integrate each term separately:

\[ \int \frac{2/5}{2x - 1} \, dx = \frac{2}{5} \cdot \frac{1}{2} \ln |2x - 1| = \frac{1}{5} \ln |2x - 1| \]

\[ \int \frac{-1/5}{x + 2} \, dx = -\frac{1}{5} \ln |x + 2| \]

Combining these results, we have:

\[ \int_{2}^{3} \left( \frac{2/5}{2x - 1} - \frac{1/5}{x + 2} \right) \, dx = \left[ \frac{1}{5} \ln |2x - 1| - \frac{1}{5} \ln |x + 2| \right]_{2}^{3} \]

### Step 4: Evaluate the Definite Integral
Now we evaluate the limits:

\[ = \frac{1}{5} \left( \ln |2(3) - 1| - \ln |3 + 2| \right) - \frac{1}{5} \left( \ln |2(2) - 1| - \ln |2 + 2| \right) \]

Calculating the values:

\[ = \frac{1}{5} \left( \ln |6 - 1| - \ln |5| \right) - \frac{1}{5} \left( \ln |4 - 1| - \ln |4| \right) \]

\[ = \frac{1}{5} \left( \ln 5 - \ln 5 \right) - \frac{1}{5} \left( \ln 3 - \ln 4 \right) \]

\[ = 0 - \frac{1}{5} \left( \ln \frac{3}{4} \right) \]

Thus, the final result is:

\[ = -\frac{1}{5} \ln \frac{3}{4} \]

### Final Answer
The value of the definite integral is:

\[ -\frac{1}{5} \ln \frac{3}{4} \]