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

Answer

To evaluate the integral

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

we will follow these steps:

### Step 1: Simplify the integrand
First, we can perform polynomial long division since the degree of the numerator is equal to the degree of the denominator. We will divide \(x^2 + x + 2\) by \((x - 2)(x - 1)\).

1. **Divide** \(x^2\) by \(x^2\) to get \(1\).
2. **Multiply** \(1\) by \((x - 2)(x - 1)\) to get \(x^2 - 3x + 2\).
3. **Subtract** this from \(x^2 + x + 2\):

\[ (x^2 + x + 2) - (x^2 - 3x + 2) = 4x. \]

Thus, we can rewrite the integral as:

\[ \int \left(1 + \frac{4x}{(x - 2)(x - 1)}\right) \, dx. \]

### Step 2: Separate the integral
Now, we can separate the integral:

\[ \int 1 \, dx + \int \frac{4x}{(x - 2)(x - 1)} \, dx. \]

### Step 3: Integrate the first part
The first integral is straightforward:

\[ \int 1 \, dx = x. \]

### Step 4: Integrate the second part using partial fractions
Next, we need to integrate \(\int \frac{4x}{(x - 2)(x - 1)} \, dx\) using partial fractions.

We can express \(\frac{4x}{(x - 2)(x - 1)}\) as:

\[ \frac{A}{x - 2} + \frac{B}{x - 1}. \]

Multiplying through by the denominator \((x - 2)(x - 1)\), we get:

\[ 4x = A(x - 1) + B(x - 2). \]

Expanding the right side:

\[ 4x = Ax - A + Bx - 2B = (A + B)x - (A + 2B). \]

### Step 5: Set up equations
Now, we can set up the equations by equating coefficients:

1. \(A + B = 4\) (coefficient of \(x\))
2. \(-A - 2B = 0\) (constant term)

From the second equation, we can express \(A\):

\[ A + 2B = 0 \implies A = -2B. \]

Substituting \(A = -2B\) into the first equation:

\[ -2B + B = 4 \implies -B = 4 \implies B = -4. \]

Then substituting \(B = -4\) back to find \(A\):

\[ A = -2(-4) = 8. \]

### Step 6: Rewrite the integral
Now we can rewrite the integral:

\[ \int \frac{4x}{(x - 2)(x - 1)} \, dx = \int \left(\frac{8}{x - 2} - \frac{4}{x - 1}\right) \, dx. \]

### Step 7: Integrate the partial fractions
Now we can integrate each term:

\[ \int \frac{8}{x - 2} \, dx - \int \frac{4}{x - 1} \, dx = 8 \ln |x - 2| - 4 \ln |x - 1|. \]

### Step 8: Combine results
Combining all parts together, we have:

\[ \int \frac{x^2 + x + 2}{(x - 2)(x - 1)} \, dx = x + 8 \ln |x - 2| - 4 \ln |x - 1| + C, \]

where \(C\) is the constant of integration.

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

\[ \int \frac{x^2 + x + 2}{(x - 2)(x - 1)} \, dx = x + 8 \ln |x - 2| - 4 \ln |x - 1| + C. \]