https://r2.erweima.ai/mathbot/image/1726441122043dd32w3eh.jpg

The integral in the image is:

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

To solve this integral, we can use **partial fraction decomposition**. Let's go through the solution step by step:

### Step 1: Factor the denominator

The quadratic term \(x^2 - 3x + 2\) can be factored. We are looking for two numbers that multiply to 2 (the constant term) and add up to -3 (the coefficient of \(x\)):

\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]

So, the integral becomes:

\[
\int \frac{2x - 1}{(x - 1)(x - 2)(2x - 3)} \, dx
\]

### Step 2: Set up partial fractions

We decompose the rational function into partial fractions as follows:

\[
\frac{2x - 1}{(x - 1)(x - 2)(2x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{2x - 3}
\]

### Step 3: Solve for \(A\), \(B\), and \(C\)

Multiply both sides by \((x - 1)(x - 2)(2x - 3)\) to clear the denominators:

\[
2x - 1 = A(x - 2)(2x - 3) + B(x - 1)(2x - 3) + C(x - 1)(x - 2)
\]

Now, expand and solve for the constants \(A\), \(B\), and \(C\).

Would you like to proceed with expanding and solving these constants, or do you prefer a summary solution?

Solve the integral: \( \int \frac{2x - 1}{(x^2 - 3x + 2)(2x - 3)} dx \)

This problem involves solving an integral using partial fraction decomposition. The quadratic equation in the denominator is factored and the integral is decomposed into simpler fractions for easy integration. Suitable for calculus students at the undergraduate or advanced high school level.

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