Evaluate : `int " "(x+(1)/(x))^(2) " dx "`

Options

A. `(x^(3))/(3)+ 2x -(1)/(x)+C`
B. `(x^(2))/(2)+ 2x -(1)/(x)+C`
C. `(x^(3))/(3)+ 2x -(1)/(x^-2)+C`
D. `(x)+ 2x -(1)/(x)+C`

Explanation

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

### Step 1: Expand the integrand
Using the identity \((a + b)^2 = a^2 + b^2 + 2ab\), we can expand the expression:
\[ (x + \frac{1}{x})^2 = x^2 + \left(\frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x} \]
This simplifies to:
\[ x^2 + \frac{1}{x^2} + 2 \]

### Step 2: Rewrite the integral
Now we can rewrite the integral as:
\[ \int (x^2 + \frac{1}{x^2} + 2) \, dx \]

### Step 3: Separate the integral
We can separate the integral into three parts:
\[ \int x^2 \, dx + \int \frac{1}{x^2} \, dx + \int 2 \, dx \]

### Step 4: Evaluate each integral
1. For \(\int x^2 \, dx\):
\[ \int x^2 \, dx = \frac{x^3}{3} \]

2. For \(\int \frac{1}{x^2} \, dx\):
\[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x} \]

3. For \(\int 2 \, dx\):
\[ \int 2 \, dx = 2x \]

### Step 5: Combine the results
Now we combine all the results from the integrals:
\[ \frac{x^3}{3} - \frac{1}{x} + 2x + C \]
where \(C\) is the constant of integration.

### Final Answer
Thus, the result of the integral is:
\[ \int (x + \frac{1}{x})^2 \, dx = \frac{x^3}{3} - \frac{1}{x} + 2x + C \]

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