`int_(0)^(1) (dx)/(sqrt(1+x)sqrt(x))=`

Options

A. `2/3 (sqrt(2)-1)`
B. `2/3(sqrt(2)+1)`
C. `4/3(sqrt(2)-1)`
D. `4/3(sqrt(2)+1)`

Answer

To solve the integral

\[ I = \int_{0}^{1} \frac{dx}{\sqrt{1+x} \sqrt{x}}, \]

we will perform a substitution and simplify the integral step by step.

### Step 1: Substitution
Let \( x = t^2 \). Then, we have:

\[ dx = 2t \, dt. \]

### Step 2: Change the limits of integration
When \( x = 0 \), \( t = 0 \). When \( x = 1 \), \( t = 1 \). Thus, the limits of integration remain from 0 to 1.

### Step 3: Substitute into the integral
Now substitute \( x \) and \( dx \) into the integral:

\[ I = \int_{0}^{1} \frac{2t \, dt}{\sqrt{1+t^2} \sqrt{t^2}}. \]

### Step 4: Simplify the integral
Since \( \sqrt{t^2} = t \) for \( t \geq 0 \), we can simplify:

\[ I = \int_{0}^{1} \frac{2t \, dt}{\sqrt{1+t^2} \cdot t} = \int_{0}^{1} \frac{2 \, dt}{\sqrt{1+t^2}}. \]

### Step 5: Factor out the constant
Now we can factor out the constant 2:

\[ I = 2 \int_{0}^{1} \frac{dt}{\sqrt{1+t^2}}. \]

### Step 6: Solve the integral
The integral \( \int \frac{dt}{\sqrt{1+t^2}} \) is a standard integral that evaluates to \( \ln(t + \sqrt{1+t^2}) + C \). Thus,

\[ \int_{0}^{1} \frac{dt}{\sqrt{1+t^2}} = \left[ \ln(t + \sqrt{1+t^2}) \right]_{0}^{1}. \]

### Step 7: Evaluate the limits
Now we evaluate the limits:

1. When \( t = 1 \):
\[ \ln(1 + \sqrt{1+1^2}) = \ln(1 + \sqrt{2}) = \ln(1 + \sqrt{2}). \]

2. When \( t = 0 \):
\[ \ln(0 + \sqrt{1+0^2}) = \ln(1) = 0. \]

Thus,

\[ \int_{0}^{1} \frac{dt}{\sqrt{1+t^2}} = \ln(1 + \sqrt{2}) - 0 = \ln(1 + \sqrt{2}). \]

### Step 8: Final result
Substituting back, we have:

\[ I = 2 \ln(1 + \sqrt{2}). \]

### Conclusion
The final answer for the integral is:

\[ \int_{0}^{1} \frac{dx}{\sqrt{1+x} \sqrt{x}} = 2 \ln(1 + \sqrt{2}). \]

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