Evaluate the following integrals:
`int frac{2x-1}{(x-1)(x+2)(x-3)}dx`

Answer

To evaluate the integral

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

we will use the method of partial fractions. Here’s the step-by-step solution:

### Step 1: Set up the partial fraction decomposition

We can express the integrand as a sum of simpler fractions:

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

where \( A \), \( B \), and \( C \) are constants to be determined.

### Step 2: Clear the denominators

Multiply both sides by the denominator \((x-1)(x+2)(x-3)\):

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

### Step 3: Expand the right-hand side

Expanding each term:

1. \( A(x+2)(x-3) = A(x^2 - 3x + 2x - 6) = A(x^2 - x - 6) \)
2. \( B(x-1)(x-3) = B(x^2 - 3x - x + 3) = B(x^2 - 4x + 3) \)
3. \( C(x-1)(x+2) = C(x^2 + 2x - x - 2) = C(x^2 + x - 2) \)

Combining these gives:

\[ 2x - 1 = (A + B + C)x^2 + (-A - 4B + C)x + (-6A + 3B - 2C) \]

### Step 4: Set up a system of equations

Now, we equate coefficients from both sides:

1. Coefficient of \( x^2 \): \( A + B + C = 0 \)
2. Coefficient of \( x \): \( -A - 4B + C = 2 \)
3. Constant term: \( -6A + 3B - 2C = -1 \)

### Step 5: Solve the system of equations

From the first equation, we have \( C = -A - B \). Substitute \( C \) into the other equations:

1. Substitute into the second equation:
\[ -A - 4B - A - B = 2 \implies -2A - 5B = 2 \implies 2A + 5B = -2 \quad (1) \]

2. Substitute into the third equation:
\[ -6A + 3B - 2(-A - B) = -1 \implies -6A + 3B + 2A + 2B = -1 \implies -4A + 5B = -1 \quad (2) \]

Now we have two equations:
1. \( 2A + 5B = -2 \) (1)
2. \( -4A + 5B = -1 \) (2)

Subtract (1) from (2):
\[ (-4A + 5B) - (2A + 5B) = -1 + 2 \implies -6A = 1 \implies A = -\frac{1}{6} \]

Substituting \( A \) back into (1):
\[ 2(-\frac{1}{6}) + 5B = -2 \implies -\frac{1}{3} + 5B = -2 \implies 5B = -2 + \frac{1}{3} = -\frac{6}{3} + \frac{1}{3} = -\frac{5}{3} \implies B = -\frac{1}{3} \]

Now substituting \( A \) and \( B \) back to find \( C \):
\[ C = -(-\frac{1}{6}) - (-\frac{1}{3}) = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]

### Step 6: Write the partial fraction decomposition

Now we have:
\[ \frac{2x-1}{(x-1)(x+2)(x-3)} = -\frac{1}{6(x-1)} - \frac{1}{3(x+2)} + \frac{1}{2(x-3)} \]

### Step 7: Integrate each term

Now we can integrate each term separately:

\[ \int \left(-\frac{1}{6(x-1)} - \frac{1}{3(x+2)} + \frac{1}{2(x-3)}\right) \, dx \]

This gives:

\[ -\frac{1}{6} \ln |x-1| - \frac{1}{3} \ln |x+2| + \frac{1}{2} \ln |x-3| + C \]

### Step 8: Combine the logarithms

We can combine the logarithmic terms:

\[ = \ln \left( \frac{(x-3)^{1/2}}{(x-1)^{1/6}(x+2)^{1/3}} \right) + C \]

### Final Answer

Thus, the final result is:

\[ \int \frac{2x-1}{(x-1)(x+2)(x-3)} \, dx = \ln \left( \frac{(x-3)^{1/2}}{(x-1)^{1/6}(x+2)^{1/3}} \right) + C \]