`int dx/sqrt(x+xsqrtx)` equals

Options

A. `logsqrt(x+sqrtxsqrtx)+c`
B. `sqrt(1+sqrtx)+c`
C. `4sqrt(1+sqrtx)+c`
D. none of these

Answer

To solve the integral \( I = \int \frac{dx}{\sqrt{x + x \sqrt{x}}} \), we can follow these steps:

### Step 1: Simplify the integrand
First, we simplify the expression inside the square root:
\[ x + x \sqrt{x} = x(1 + \sqrt{x}) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{\sqrt{x(1 + \sqrt{x})}} \]

### Step 2: Factor out \(\sqrt{x}\)
Next, we can factor out \(\sqrt{x}\) from the square root:
\[ \sqrt{x(1 + \sqrt{x})} = \sqrt{x} \cdot \sqrt{1 + \sqrt{x}} \]
This gives us:
\[ I = \int \frac{dx}{\sqrt{x} \sqrt{1 + \sqrt{x}}} \]

### Step 3: Substitute \( t = 1 + \sqrt{x} \)
Now, let's make the substitution \( t = 1 + \sqrt{x} \). Then, we have:
\[ \sqrt{x} = t - 1 \quad \text{and} \quad x = (t - 1)^2 \]
Differentiating both sides gives:
\[ dx = 2(t - 1) dt \]

### Step 4: Substitute in the integral
Substituting \( \sqrt{x} \) and \( dx \) into the integral:
\[ I = \int \frac{2(t - 1) dt}{(t - 1) \sqrt{1 + \sqrt{x}}} \]
Since \( \sqrt{1 + \sqrt{x}} = \sqrt{t} \), we can rewrite the integral:
\[ I = \int \frac{2 dt}{\sqrt{t}} \]

### Step 5: Integrate
Now, we can integrate:
\[ I = 2 \int t^{-1/2} dt = 2 \cdot 2 t^{1/2} + C = 4 \sqrt{t} + C \]

### Step 6: Substitute back for \( t \)
Substituting back for \( t \):
\[ I = 4 \sqrt{1 + \sqrt{x}} + C \]

Thus, the final result is:
\[ \int \frac{dx}{\sqrt{x + x \sqrt{x}}} = 4 \sqrt{1 + \sqrt{x}} + C \]