`int(sqrt(1+sqrtx))/(sqrtx)dx=`

Options

A. `sqrt(logx+2sqrtx)+c`
B. `(4)/(3)(1+sqrtx)^(3//2)+c`
C. `(1)/(sqrt(logx+2sqrtx))+c`
D. `(4)/(3(1+sqrtx)^(3//2))+c`

Answer

To solve the integral \( \int \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{x}} \, dx \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we have:
\[ x = t^2 \quad \text{and} \quad dx = 2t \, dt \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we get:
\[ \int \frac{\sqrt{1 + t}}{t} \cdot 2t \, dt = 2 \int \sqrt{1 + t} \, dt \]

### Step 3: Integrate
Now we need to integrate \( \sqrt{1 + t} \):
\[ \int \sqrt{1 + t} \, dt \]
Using the power rule for integration, we can rewrite \( \sqrt{1 + t} \) as \( (1 + t)^{1/2} \):
\[ \int (1 + t)^{1/2} \, dt = \frac{(1 + t)^{3/2}}{\frac{3}{2}} + C = \frac{2}{3} (1 + t)^{3/2} + C \]

### Step 4: Substitute Back
Now we substitute back \( t = \sqrt{x} \):
\[ 2 \left( \frac{2}{3} (1 + t)^{3/2} + C \right) = \frac{4}{3} (1 + \sqrt{x})^{3/2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{x}} \, dx = \frac{4}{3} (1 + \sqrt{x})^{3/2} + C \]

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