`int((x^(3)+1)^(2))/(x^(2))dx=`

Options

A. `(x^(5))/(5)-x^(2)+(1)/(x)+c`
B. `x^(2)-(x^(5))/(5)+x+c`
C. `(x^(5))/(5)+x^(2)-(1)/(x)+c`
D. `tan^(-1)(x^(2))+c`

Answer

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

### Step 1: Expand the integrand
First, we need to expand the expression \((x^3 + 1)^2\):
\[ (x^3 + 1)^2 = x^6 + 2x^3 + 1 \]
Thus, we can rewrite the integral as:
\[ \int \frac{x^6 + 2x^3 + 1}{x^2} \, dx \]

### Step 2: Simplify the expression
Now, we can separate the terms in the integrand:
\[ \int \left( \frac{x^6}{x^2} + \frac{2x^3}{x^2} + \frac{1}{x^2} \right) \, dx = \int (x^4 + 2x + x^{-2}) \, dx \]

### Step 3: Integrate each term
Now we can integrate each term separately:
1. \(\int x^4 \, dx = \frac{x^5}{5}\)
2. \(\int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2\)
3. \(\int x^{-2} \, dx = -\frac{1}{x}\)

Putting it all together, we have:
\[ \int (x^4 + 2x + x^{-2}) \, dx = \frac{x^5}{5} + x^2 - \frac{1}{x} + C \]

### Final Answer
Thus, the final answer is:
\[ \frac{x^5}{5} + x^2 - \frac{1}{x} + C \]