`int((x-2)^3)/x^2dx`=?

Answer

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

### Step 1: Expand the integrand
We start by expanding the expression \((x-2)^3\) using the binomial expansion formula:
\[ (x-2)^3 = x^3 - 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2 - 2^3 \]
This simplifies to:
\[ x^3 - 6x^2 + 12x - 8 \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{(x-2)^3}{x^2} \, dx = \int \frac{x^3 - 6x^2 + 12x - 8}{x^2} \, dx \]
This can be separated into individual terms:
\[ \int \left( \frac{x^3}{x^2} - \frac{6x^2}{x^2} + \frac{12x}{x^2} - \frac{8}{x^2} \right) \, dx = \int (x - 6 + \frac{12}{x} - \frac{8}{x^2}) \, dx \]

### Step 3: Integrate each term
Now we can integrate each term separately:
1. \(\int x \, dx = \frac{x^2}{2}\)
2. \(\int -6 \, dx = -6x\)
3. \(\int \frac{12}{x} \, dx = 12 \ln |x|\)
4. \(\int -\frac{8}{x^2} \, dx = 8 \cdot \frac{1}{x}\)

Putting it all together, we have:
\[ \int (x - 6 + \frac{12}{x} - \frac{8}{x^2}) \, dx = \frac{x^2}{2} - 6x + 12 \ln |x| + 8 \cdot \frac{1}{x} + C \]

### Step 4: Combine the results
Thus, the final result of the integral is:
\[ \frac{x^2}{2} - 6x + 12 \ln |x| + \frac{8}{x} + C \]

### Final Answer:
\[ \int \frac{(x-2)^3}{x^2} \, dx = \frac{x^2}{2} - 6x + 12 \ln |x| + \frac{8}{x} + C \]

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