`int(sqrtx+1/sqrtx)^2 dx=`

Options

A. `x^2/2 log x-2x +c`
B. `x^2/2 +log x+2x +c`
C. `x^2/2 - log x+2x+c`
D. `x^2/2 +logx+4x+c`

Answer

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

### Step 1: Expand the integrand
First, we expand the expression inside the integral:

\[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = \left(\sqrt{x}\right)^2 + 2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2 \]

This simplifies to:

\[ x + 2 + \frac{1}{x} \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:

\[ \int \left(x + 2 + \frac{1}{x}\right) \, dx \]

### Step 3: Integrate term by term
Now we will integrate each term separately:

1. \(\int x \, dx = \frac{x^2}{2}\)
2. \(\int 2 \, dx = 2x\)
3. \(\int \frac{1}{x} \, dx = \ln |x|\)

Putting it all together, we have:

\[ \int \left(x + 2 + \frac{1}{x}\right) \, dx = \frac{x^2}{2} + 2x + \ln |x| + C \]

where \(C\) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:

\[ \int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 \, dx = \frac{x^2}{2} + 2x + \ln |x| + C \]

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