`int((x-4)^2)/x^3dx`=?

Explanation

To solve the integral \(\int \frac{(x-4)^2}{x^3} \, dx\), we will follow these steps:

### Step 1: Expand the numerator
We start by expanding the expression \((x-4)^2\):
\[ (x-4)^2 = x^2 - 8x + 16 \]
Thus, we can rewrite the integral as:
\[ \int \frac{x^2 - 8x + 16}{x^3} \, dx \]

### Step 2: Split the integral
We can split the integral into three separate integrals:
\[ \int \left( \frac{x^2}{x^3} - \frac{8x}{x^3} + \frac{16}{x^3} \right) \, dx = \int \left( \frac{1}{x} - \frac{8}{x^2} + \frac{16}{x^3} \right) \, dx \]

### Step 3: Integrate each term
Now we will integrate each term separately:
1. \(\int \frac{1}{x} \, dx = \ln |x|\)
2. \(\int -\frac{8}{x^2} \, dx = -8 \int x^{-2} \, dx = -8 \left( -\frac{1}{x} \right) = \frac{8}{x}\)
3. \(\int \frac{16}{x^3} \, dx = 16 \int x^{-3} \, dx = 16 \left( -\frac{1}{2x^2} \right) = -\frac{8}{x^2}\)

### Step 4: Combine the results
Now we combine the results of the integrals:
\[ \ln |x| + \frac{8}{x} - \frac{8}{x^2} + C \]
where \(C\) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{(x-4)^2}{x^3} \, dx = \ln |x| + \frac{8}{x} - \frac{8}{x^2} + C \]

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