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 a step-by-step solution:

### Step 1: Set Up Partial Fractions
We can express the integrand as a sum of partial 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 that we need to determine.

### Step 2: Clear the Denominator
Multiply both sides by \((x-1)(x+2)(x-3)\):

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

### Step 3: Expand the Right Side
Now, we will expand the right-hand side:

1. For \(A(x+2)(x-3)\):
\[ A(x^2 - 3x + 2x - 6) = A(x^2 - x - 6) \]

2. For \(B(x-1)(x-3)\):
\[ B(x^2 - 3x - x + 3) = B(x^2 - 4x + 3) \]

3. For \(C(x-1)(x+2)\):
\[ C(x^2 + 2x - x - 2) = C(x^2 + x - 2) \]

Combining these gives:

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

### Step 4: Set Up 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
We have the following system of equations:

1. \(A + B + C = 0\)
2. \(-A - 4B + C = 2\)
3. \(-6A + 3B - 2C = -1\)

From equation 1, we can express \(C\) in terms of \(A\) and \(B\):

\[ C = -A - B \]

Substituting \(C\) into the other two equations:

1. \(-A - 4B - A - B = 2 \Rightarrow -2A - 5B = 2\)
2. \(-6A + 3B - 2(-A - B) = -1 \Rightarrow -6A + 3B + 2A + 2B = -1 \Rightarrow -4A + 5B = -1\)

Now we solve these two equations:

1. \(2A + 5B = -2\) (Equation 4)
2. \(-4A + 5B = -1\) (Equation 5)

Subtract Equation 4 from Equation 5:

\[ (-4A + 5B) - (2A + 5B) = -1 + 2 \]
\[ -6A = 1 \Rightarrow A = -\frac{1}{6} \]

Substituting \(A\) back into Equation 4:

\[ 2(-\frac{1}{6}) + 5B = -2 \Rightarrow -\frac{1}{3} + 5B = -2 \Rightarrow 5B = -2 + \frac{1}{3} = -\frac{6}{3} + \frac{1}{3} = -\frac{5}{3} \Rightarrow B = -\frac{1}{3} \]

Now substituting \(A\) and \(B\) back into \(C = -A - B\):

\[ C = -(-\frac{1}{6}) - (-\frac{1}{3}) = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]

### Step 6: Write the Partial Fraction Decomposition
Now we have:

\[ A = -\frac{1}{6}, \quad B = -\frac{1}{3}, \quad C = \frac{1}{2} \]

Thus, we can write:

\[ \frac{2x-1}{(x-1)(x+2)(x-3)} = -\frac{1}{6(x-1)} - \frac{1}{3(x+2)} + \frac{1/2}{x-3} \]

### Step 7: Integrate Each Term
Now we can integrate term by term:

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

This gives:

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

### Final Answer
Thus, the final result of the integral is:

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