`int((x+1)(2x-3))/(x) dx`

Answer

To solve the integral \(\int \frac{(x+1)(2x-3)}{x} \, dx\), we can follow these steps:

### Step 1: Expand the numerator
First, we need to expand the expression \((x+1)(2x-3)\):

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

### Step 2: Rewrite the integral
Now, we can rewrite the integral:

\[ \int \frac{2x^2 - x - 3}{x} \, dx \]

### Step 3: Simplify the integrand
Next, we can simplify the integrand by dividing each term by \(x\):

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

### Step 4: Integrate term by term
Now we can integrate each term separately:

1. \(\int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2\)
2. \(\int -1 \, dx = -x\)
3. \(\int -\frac{3}{x} \, dx = -3 \ln |x|\)

Combining these results, we have:

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

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

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