Evaluate the following :
`int (x-4)^3/x^2 dx.`

Answer

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

### Step 1: Expand the integrand
We start by expanding \( (x-4)^3 \) using the binomial expansion formula:
\[ (x-4)^3 = x^3 - 3(4)x^2 + 3(4^2)x - 4^3 \]
Calculating each term:
- \( 4^2 = 16 \)
- \( 4^3 = 64 \)

Thus, we have:
\[ (x-4)^3 = x^3 - 12x^2 + 48x - 64 \]

### Step 2: Rewrite the integral
Now, substitute the expanded form back into the integral:
\[ \int \frac{(x-4)^3}{x^2} \, dx = \int \frac{x^3 - 12x^2 + 48x - 64}{x^2} \, dx \]
This can be simplified by dividing each term by \( x^2 \):
\[ = \int \left( x - 12 + \frac{48}{x} - \frac{64}{x^2} \right) \, dx \]

### Step 3: Integrate term by term
Now we can integrate each term separately:
1. The integral of \( x \) is \( \frac{x^2}{2} \).
2. The integral of \( -12 \) is \( -12x \).
3. The integral of \( \frac{48}{x} \) is \( 48 \ln |x| \).
4. The integral of \( -\frac{64}{x^2} \) is \( \frac{64}{x} \) (since the integral of \( x^{-2} \) is \( -x^{-1} \)).

Putting it all together, we have:
\[ \int \left( x - 12 + \frac{48}{x} - \frac{64}{x^2} \right) \, dx = \frac{x^2}{2} - 12x + 48 \ln |x| + \frac{64}{x} + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{(x-4)^3}{x^2} \, dx = \frac{x^2}{2} - 12x + 48 \ln |x| + \frac{64}{x} + C \]

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