The antiderivative of `(sqrtx+1/sqrtx)` equals:

Options

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

Answer

To find the antiderivative of the function \( \sqrt{x} + \frac{1}{\sqrt{x}} \), we will follow these steps:

### Step 1: Rewrite the Function
First, we rewrite \( \sqrt{x} \) and \( \frac{1}{\sqrt{x}} \) in terms of exponents:
\[ \sqrt{x} = x^{1/2} \quad \text{and} \quad \frac{1}{\sqrt{x}} = x^{-1/2} \]
Thus, the function can be expressed as:
\[ x^{1/2} + x^{-1/2} \]

### Step 2: Set Up the Integral
Now, we need to find the integral of the function:
\[ \int \left( x^{1/2} + x^{-1/2} \right) \, dx \]

### Step 3: Integrate Each Term
We can integrate each term separately using the power rule for integration, which states:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

Applying this to each term:
1. For \( x^{1/2} \):
\[ \int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]

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

### Step 4: Combine the Results
Now, we combine the results of the integrals:
\[ \int \left( x^{1/2} + x^{-1/2} \right) \, dx = \frac{2}{3} x^{3/2} + 2 x^{1/2} + C \]

### Final Answer
Thus, the antiderivative of \( \sqrt{x} + \frac{1}{\sqrt{x}} \) is:
\[ \frac{2}{3} x^{3/2} + 2 x^{1/2} + C \]

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