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

Answer

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

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

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

### Step 2: Rewrite the Integral
Now we can rewrite the integral using the expanded numerator:

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

### Step 3: Simplify the Fraction
Next, we simplify the fraction by dividing each term in the numerator 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: Separate the Integral
Now we can separate the integral into three simpler integrals:

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

### Step 5: Integrate Each Term
Now we will integrate each term separately:

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

Putting it all together, we have:

\[ x^2 - x - 3 \ln |x| + C \]

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

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

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